Sifat Turunan Pertama dan Aturan Rantai pada Turunan Fungsi Aljabar

Rumus Turunan dan Sifat Turunan Pertama

$\begin{array}{rl}& \\ \text{Untuk}:& \\ & \{\begin{array}{ll}a& \in \mathbb{R}\\ n& \in \mathbb{Q}\\ c& \text{konstanta}\\ U& =g(x)\\ V& =h(x)\end{array}\\ & \end{array}$.

$\begin{array}{|l|}\hline \mathbf{\text{Sifat-Sifat}}\\ \begin{array}{rl}y& =c\to y^{\prime }=0\\ y& =c.U\to y^{\prime }=c.U^{\prime }\\ y& =U\pm V\to y^{\prime }=U^{\prime }\pm V^{\prime }\\ y& =U.V\to y^{\prime }=U^{\prime }.V+U.V^{\prime }\\ y& ={\displaystyle \frac{U}{V}\to y^{\prime }={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}}\end{array}\\ \mathbf{\text{Fungsi Aljabar}}\\ \begin{array}{rl}y& =a.x^n\to y{}^{\prime }=n.a.x^{(n-1)}\\ y& =a.U^n\to y{}^{\prime }=n.a.U^{(n-1)}.U^{\prime }\end{array}\\ \mathbf{\text{Fungsi Trigonometri}}\\ \begin{array}{rlrl}y& =a\sin U\to & y^{\prime }& =(a\cos U).U^{\prime }\\ y& =a\cos U\to & y^{\prime }& =(-a\sin U).U^{\prime }\\ y& =a\tan U\to & y^{\prime }& =(a{\sec }^{2}U).U^{\prime }\end{array}\\ \mathbf{\text{Aturan rantai }}\\ \begin{array}{rl}& \text{pada turunan untuk}y=f(u),\text{jika}\\ & \text{untuk}u\text{merupakan fungsi}x,\text{maka}:\\ & y{}^{\prime }=f{}^{\prime }(x).u^{\prime }\text{atau}{\displaystyle \frac{dy}{dx}={\displaystyle \frac{dy}{du}.\frac{du}{dx}}}\\ & \end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Dengan menggunakan rumus turunan}\\ & f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h},}\\ & \text{tunjukkan bahwa turunan}\\ & \text{a}.f(x)=a{x}^n\text{adalah}f{}^{\prime }(x)=n.a{x}^{n-1}\\ & \text{b}.f(x)=u(x)+v(x)\text{adalah}f{}^{\prime }(x)=u{}^{\prime }(x)+v{}^{\prime }(x)\\ & \text{c}.f(x)=u(x).v(x)\text{adalah}f{}^{\prime }(x)=u{}^{\prime }(x).v(x)+u(x).v{}^{\prime }(x)\\ & \text{d}.f(x)={\displaystyle \frac{u(x)}{v(x)}\text{adalah}f{}^{\prime }(x)={\displaystyle \frac{u{}^{\prime }(x).v(x)+u(x).v{}^{\prime }(x)}{(v(x){)}^2}}}\\ & \text{e}.f(x)=\sin x\text{adalah}f{}^{\prime }(x)=\cos x\\ & \text{f}.f(x)=\cos x\text{adalah}f{}^{\prime }(x)=-\sin x\\ & \text{g}.f(x)=\tan x\text{adalah}f{}^{\prime }(x)={\sec }^{2}x\\ & \text{h}.f(x)=\cot x\text{adalah}f{}^{\prime }(x)=-{\csc }^{2}x\end{array}$.

$\begin{array}{rl}& \mathbf{\text{Bukti}}\\ & \begin{array}{l}\\ 1.\text{a}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a(x+h{)}^n-a{x}^n}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a(x^n+(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}.h+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}.h^2+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}.h^3+\cdots +(\genfrac{}{}{0ex}{}{n}{n-1})x.h^{n-1}+h^n)-a{x}^n}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a((\genfrac{}{}{0ex}{}{n}{1})x^{n-1}.h+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}.h^2+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}.h^3+\cdots +(\genfrac{}{}{0ex}{}{n}{n-1})x.h^{n-1}+h^n)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah((\genfrac{}{}{0ex}{}{n}{1})x^{n-1}+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}.h+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}.h^2+\cdots +(\genfrac{}{}{0ex}{}{n}{n-1})x.h^{n-2}+h^{n-1})}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle a(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}+a(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}.h+a(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}.h^2+\cdots +a{h}^{n-1}}\\ & =a(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}+0+0+...+0\\ & =a.{\displaystyle \frac{n!}{(n-1)!.1!}{x}^{n-1}}\\ & =a.{\displaystyle \frac{n.(n-1)!}{(n-1)!}{x}^{n-1}}\\ & =a.n.x^{n-1}◼\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{b}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(u(x+h)+v(x+h))-(u(x)+v(x))}{h}}\\ & =\underset{h\to 0}{\text{Lim}}\left({\displaystyle \frac{u(x+h)-u(x)}{h}+\frac{v(x+h)-v(x)}{h}}\right)\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{u(x+h)-u(x)}{h}+\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{v(x+h)-v(x)}{h}}}\\ & =u^{\prime }(x)+v^{\prime }(x)◼\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{c}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(u(x+h)\times v(x+h))-(u(x)\times v(x))}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{u(x+h)\times v(x+h)-u(x+h)\times v(x)+u(x+h)\times v(x)-u(x)\times v(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}(u(x+h)\times {\displaystyle \frac{v(x+h)-v(x)}{h}+v(x)\times {\displaystyle \frac{u(x+h)-u(x)}{h}}})\\ & =\underset{h\to 0}{\text{Lim}}u(x+h)\times \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{v(x+h)-v(x)}{h}+\underset{h\to 0}{\text{Lim}}v(x)\times \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{u(x+h)-u(x)}{h}}}\\ & =u(x)\times v^{\prime }(x)+v(x)\times u^{\prime }(x)\\ & =u^{\prime }(x)\times v(x)+u(x)\times v^{\prime }(x)◼\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{d}& \text{Misalkan}p(x)={\displaystyle \frac{u(x)}{v(x)}}\\ & \text{Sebelumnya telah diketahui dari no. 1. c}\\ & u(x)=p(x)\times v(x)\\ & \begin{array}{rl}u{}^{\prime }(x)& =p^{\prime }(x)\times v(x)+p(x)\times v^{\prime }(x)\\ \text{Sekar}& \text{ang kita substitusikan pemisalan}\\ \text{di at}& \text{as, yaitu}:\end{array}\\ & \begin{array}{rl}{p}^{\prime }(x)& \times v(x)=u^{\prime }(x)-p(x)\times v^{\prime }(x)\\ & =u^{\prime }(x)-{\displaystyle \frac{u(x)}{v(x)}\times v^{\prime }(x)}\\ & ={\displaystyle \frac{u^{\prime }(x)\times v(x)-u(x)\times v^{\prime }(x)}{v(x)}}\\ p^{\prime }(x)& ={\displaystyle \frac{u^{\prime }(x)\times v(x)-u(x)\times v^{\prime }(x)}{v^2(x)}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{e}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)-\sin x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}(2x+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\cos {\displaystyle \frac{1}{2}(2x+h).{\displaystyle \frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 2\cos {\displaystyle \frac{1}{2}(2x+h)\times {\displaystyle \frac{1}{2}}}}\\ & =2\cos {\displaystyle \frac{1}{2}(2x+0)\times {\displaystyle \frac{1}{2}}}\\ & =\cos {\displaystyle \frac{1}{2}(2x)}\\ & =\cos x◼\end{array}\end{array}$ .

$\begin{array}{l}\\ 1.\text{f}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\cos (x+h)-\cos x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2\sin {\displaystyle \frac{1}{2}(2x+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle -2\sin {\displaystyle \frac{1}{2}(2x+h).\frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle -2\sin {\displaystyle \frac{1}{2}(2x+h)\times \frac{1}{2}}}\\ & =-2\sin {\displaystyle \frac{1}{2}(2x+0)\times \frac{1}{2}}\\ & =-\sin {\displaystyle \frac{1}{2}(2x)}\\ & =-\sin x◼\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{g}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan (x+h)-\tan x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\tan x+\tan h}{1-\tan x.\tan h}-\tan x}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\tan x+\tan h-\tan x+{\tan }^{2}x.\tan h}{1-\tan x.\tan h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan h(1+{\tan }^{2}x)}{h(1-\tan x.\tan h)}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan h}{h}\times \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{1+{\tan }^{2}x}{1-\tan x.\tan h}}}\\ & =1\times {\displaystyle \frac{1+{\tan }^{2}x}{1-0}}\\ & =1+{\tan }^{2}x\\ & ={\sec }^{2}x◼\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah turunannya}\\ & \begin{array}{l}\\ \text{a}.& f(x)=x^6& \text{i}.& f(x)=(x+1)(x-2)& \text{q}.& f(x)={\displaystyle \frac{2}{x^3}}\\ \text{b}.& f(x)={\displaystyle \frac{1}{2}{x}^2}& \text{j}.& f(x)=(3-x)(5-x)& \text{r}.& f(x)={\displaystyle \frac{1}{2\sqrt{x}}}\\ \text{c}.& f(x)=-2x^5& \text{k}.& f(x)=(2x+3{)}^2& \text{s}.& f(x)={\displaystyle \frac{1}{3x^5}}\\ \text{d}.& f(x)=a{x}^3& \text{l}.& f(x)=(x-2{)}^3& \text{t}.& f(x)=2x^4+{\displaystyle \frac{1}{2x^3}}\\ \text{e}.& f(x)=4x^4-x^2+2017& \text{m}.& f(x)=(4x+1)(4x-1)& \text{u}.& f(x)={\displaystyle \frac{x}{2}+\frac{2}{x}}\\ \text{f}.& f(x)=6-x-3x^2& \text{n}.& f(x)=(x-1)(x+1)(x+2)& \text{v}.& f(x)={(2x^3+{\displaystyle \frac{1}{x}})}^2\\ \text{g}.& f(x)=(x-3{)}^2& \text{o}.& f(x)={\displaystyle \frac{1}{2}{x}^{-4}}& \text{w}.& f(x)=x^2{(1+\sqrt{x})}^2\\ \text{h}.& f(x)={(x^3-2)}^2& \text{p}.& f(x)=x^{-5}& \text{x}.& f(x)={\displaystyle \frac{2+4x}{x^2}}\\ & & & & \text{y}.& f(x)=(x+1+{\displaystyle \frac{1}{x}})(x+1-{\displaystyle \frac{1}{x}})\end{array}\end{array}$.

$\begin{array}{rl}& \text{Untuk}y=f(x)\text{maka}\\ & \begin{array}{|ccc|}\hline \text{a}& \text{b}& \text{c}\\ \begin{array}{rl}y& =x^6\\ y{}^{\prime }& =6x^{6-1}\\ & =6x^5\\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{1}{2}{x}^2}\\ y{}^{\prime }& =2.{\displaystyle \frac{1}{2}{x}^{2-1}}\\ & =x\end{array}& \begin{array}{rl}y& =-2x^5\\ y{}^{\prime }& =-5.2x^{5-1}\\ & =-10x^4\\ & \end{array}\\ \text{d}& \text{e}& \text{f}\\ \begin{array}{rl}y& =a{x}^3\\ y{}^{\prime }& =3.a.x^{3-1}\\ & =3.a.x^2\end{array}& \begin{array}{rl}y& =4x^4-x^2+2017\\ y{}^{\prime }& =4.4x^{4-1}-2.x^{2-1}+0\\ & =16x^3-2x\end{array}& \begin{array}{rl}y& =6-x-3x^2\\ y{}^{\prime }& =0-1.x^{1-1}-2.3x^{2-1}\\ & =-1-6x\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{|cc|}\hline \text{g}& \text{h}\\ \begin{array}{rl}y& =(x-3{)}^2\\ y{}^{\prime }& =2.(x-3{)}^{2-1}.1\\ & =2(x-3)\\ & \end{array}& \begin{array}{rl}y& ={(x^3-2)}^2\\ y{}^{\prime }& =2.{(x^3-2)}^{2-1}.3x^2\\ & =6(x^3-2)x^2\\ & =6x^5-12x^2\end{array}\\ \text{i}& \text{j}\\ \begin{array}{rl}y& =(x+1)(x-2)\\ y{}^{\prime }& =1.(x+2)+(x+1).1\\ & =2x+3\end{array}& \begin{array}{rl}y& =(3-x)(5-x)\\ y{}^{\prime }& =-1.(5-x)+(3-x).-1\\ & =2x-8\end{array}\\ \hline\end{array}$ .

$\begin{array}{|cccc|}\hline \text{k}& \text{l}& {\text{m}}_1& {\text{m}}_2\\ \begin{array}{rl}y& =(2x+3{)}^2\\ y{}^{\prime }& =2.(2x+3{)}^{2-1}.2\\ & =4(2x+3{)}^1\\ & =8x+12\\ & \\ & \end{array}& \begin{array}{rl}y& =(x-2{)}^3\\ y{}^{\prime }& =3(x-2{)}^{3-1}.1\\ & =3(x-2{)}^2\\ & \\ & \\ & \end{array}& \begin{array}{rl}y& =(4x+1)(4x-1)\\ \text{c}& \text{ara 1}\\ y{}^{\prime }& =4(4x-1)+(4x+1).4\\ & =16x-4+16x+4\\ & =32x\\ & \end{array}& \begin{array}{rl}y& =(4x+1)(4x-1)\\ \text{c}& \text{ara 2}\\ y& =16x^2-1\\ y{}^{\prime }& =2.16x^{2-1}-0\\ & =32x^1\\ & =32x\end{array}\\ \text{n}& \text{o}& \text{p}& \text{q}\\ \begin{array}{rl}y& =(x-1)(x+1)(x+2)\\ & =(x^2-1)(x+2)\\ & =x^3+2x^2-x-2\\ y{}^{\prime }& =3x^{3-1}+2.2x^{2-1}-x^{1-1}-0\\ & =3x^2+4x-1\\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{1}{2}{x}^{-4}}\\ y{}^{\prime }& =-4.{\displaystyle \frac{1}{2}{x}^{-4-1}}\\ & =-2x^{-5}\\ & =-{\displaystyle \frac{2}{x^5}}\\ & \end{array}& \begin{array}{rl}y& =x^{-5}\\ y^{\prime }& =-5.x^{-5-1}\\ & =-5x^{-6}\\ & =-{\displaystyle \frac{5}{x^6}}\\ & \\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{2}{x^3}}\\ & =2x^{-3}\\ y{}^{\prime }& =-3.2x^{-3-1}\\ & =-6x^{-4}\\ & =-{\displaystyle \frac{6}{x^4}}\end{array}\\ \hline\end{array}$.

$\begin{array}{|cccc|}\hline \text{r}& \text{s}& \text{t}& \text{u}\\ \begin{array}{rl}y& ={\displaystyle \frac{1}{2\sqrt{x}}}\\ & ={\displaystyle \frac{1}{2x^{\frac{1}{2}}}}\\ & ={\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}\\ y{}^{\prime }& =-{\displaystyle \frac{1}{2}.\frac{1}{2}{x}^{{}^{-\frac{1}{2}-1}}}\\ & =-{\displaystyle \frac{1}{4}{x}^{{}^{-\frac{3}{2}}}}\\ & =-{\displaystyle \frac{1}{4x^{{}^{\frac{3}{2}}}}}\\ & =-{\displaystyle \frac{1}{4}\sqrt{x^3}}\end{array}& \begin{array}{rl}y& ={\displaystyle \frac{1}{3x^5}}\\ & ={\displaystyle \frac{1}{3}{x}^{-5}}\\ y{}^{\prime }& =-5.{\displaystyle \frac{1}{3}{x}^{-5-1}}\\ & =-{\displaystyle \frac{5}{3}{x}^{-6}}\\ & =-{\displaystyle \frac{5}{3x^6}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}y& =2x^4+{\displaystyle \frac{1}{2x^3}}\\ & =2x^4+{\displaystyle \frac{1}{2}{x}^{-3}}\\ y{}^{\prime }& =4.2x^{4-1}+(-3).{\displaystyle \frac{1}{2}{x}^{-3-1}}\\ & =8x^3-{\displaystyle \frac{3}{2}{x}^{-4}}\\ & =8x^3-{\displaystyle \frac{3}{2x^4}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{x}{2}+\frac{2}{x}}\\ & ={\displaystyle \frac{1}{2}x+2x^{-1}}\\ y{}^{\prime }& ={\displaystyle \frac{1}{2}{x}^{1-1}+(-1).2x^{-1-1}}\\ & ={\displaystyle \frac{1}{2}{x}^0-2x^{-2}}\\ & ={\displaystyle \frac{1}{2}-{\displaystyle \frac{2}{x^2}}}\\ & \\ & \\ & \end{array}\\ \hline\end{array}$.

$\begin{array}{|cc|}\hline \text{v}& \text{w}\\ \begin{array}{rl}y& ={(2x^3+{\displaystyle \frac{1}{x}})}^2\\ & ={(2x^3+x^{-1})}^2\\ y{}^{\prime }& =2.{(2x^3+x^{-1})}^{2-1}.(3.2x^{3-1}+(-1)x^{-1-1})\\ & =2.{(2x^3+x^{-1})}^1.(6x^2-x^{-2})\\ & =2(2x^3+{\displaystyle \frac{1}{x}})(6x^2-{\displaystyle \frac{1}{x^2}})\\ & =2(12x^5-2x+6x-{\displaystyle \frac{1}{x^3}})\\ & =24x^5+8x-{\displaystyle \frac{2}{x^3}}\\ & \end{array}& \begin{array}{rl}y& =x^2{(1+\sqrt{x})}^2\\ & =x^2{(1+x^{{}^{\frac{1}{2}}})}^2\\ y{}^{\prime }& =2x.{(1+x^{{}^{\frac{1}{2}}})}^2+x^2.2.{(1+x^{{}^{\frac{1}{2}}})}^{2-1}.(0+{\displaystyle \frac{1}{2}.x^{{}^{\frac{1}{2}-1}}})\\ & =2x{(1+\sqrt{x})}^2+2x^2.(1+\sqrt{x}).\left({\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}\right)\\ & =2x{(1+\sqrt{x})}^2+2x^2.\left({\displaystyle \frac{1}{2x^{{}^{\frac{1}{2}}}}}\right).(1+\sqrt{x})\\ & =2x{(1+\sqrt{x})}^2+x^{{}^{2-\frac{1}{2}}}.(1+\sqrt{x})\\ & =2x{(1+\sqrt{x})}^2+x^{{}^{\frac{3}{2}}}(1+\sqrt{x})\\ & =2x{(1+\sqrt{x})}^2+(x\sqrt{x}+x^2)\end{array}\\ \hline\end{array}$.

$\begin{array}{|cc|}\hline {\text{x}}_1& {\text{x}}_2\\ \begin{array}{rl}y& ={\displaystyle \frac{U}{V}}\\ y{}^{\prime }& ={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}\end{array}& \begin{array}{rl}y& =UV\\ y{}^{\prime }& =U^{\prime }.V+U.V^{\prime }\end{array}\\ \begin{array}{rlrl}& & & \\ U& =2+4x& \to U^{\prime }& =4\\ V& =x^2& \to V^{\prime }& =2x\\ & & & \end{array}& \begin{array}{rlrl}& & & \\ U& =2+4x& \to U^{\prime }& =4\\ V& =x^2& \to V^{\prime }& =2x\\ & & & \end{array}\\ \begin{array}{rl}y& ={\displaystyle \frac{2+4x}{x^2}}\\ y{}^{\prime }& ={\displaystyle \frac{(4)(x^2)-(2+4x).(2x)}{{\left(x^2\right)}^2}}\\ & ={\displaystyle \frac{4x^2-4x-8x^2}{x^4}}\\ & ={\displaystyle \frac{-4x^2-4x}{x^4}}\\ & ={\displaystyle \frac{-4x-4}{x^3}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{2+4x}{x^2}}\\ & =(2+4x).x^{-2}\\ y{}^{\prime }& =(4).x^{-2}+(2+4x).-2x^{-2-1}\\ & =4x^{-2}-(4+8x).x^{-3}\\ & =4x^{-2}-4x^{-3}-8x^{-2}\\ & =-4x^{-3}-4x^{-2}\\ & ={\displaystyle \frac{-4}{x^3}+{\displaystyle \frac{-4}{x^2}}}\\ & ={\displaystyle \frac{-4-4x}{x^3}}\\ & ={\displaystyle \frac{-4x-4}{x^3}}\end{array}\\ \hline\end{array}$.

Pengertian dan Bentuk Umum Turunan Fungsi Aljabar (Materi Lanjutan Turunan Fungsi Aljabar)

A. 2 Pengertian Turunan Fungsi Aljabar

Perhatikan ilustrasi gambar berikut. 

Misalkan diketahui fungsi  $y=f(x)$  terdefinisi pada semua nilai  $x$ di sekitar   $x=k$. Jika  $\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$  ada, maka bentuk  $\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$  disebut sebagai turunan dari fungsi  $f(x)$  saat  $x=k$.

A. 3 Notasi

• Notasi turunan fungsi dilambangkan dengan  $f^{\prime }(k)$  dengan  $f^{\prime }(k)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$.
• Lambang   $f^{\prime }(k)$  dibaca   $f$  aksen   $k$ disebut turunan atau derivatif untuk fungsi   $f(x)$ terhadap   $x$  saat   $x=k$.
• Jika limitnya ada, dapat dikatakan fungsi   $f(x)$ diferensiabel (dapat dideferensialkan) saat   $x=k$  dan bentuk limitnya selanjutnya dilambangkan dengan  $f^{\prime }(k)$.
• Misalkan fungsi  $f(x)$  mempunyai turunan  $f^{\prime }(x)$. Jika  $f^{\prime }(k)$  tidak terdefinisi, maka  $f(x)$  tidak diferensiabel di  $x=k$.

A. 4 Bentuk Umum Turunan Pertama Fungsi Aljabar

Bentuk umum turunan pertama fungsi aljabar  untuk fungsi  $y$ terhadap $x$  dinotasikan sebagaimana berikut

$y{}^{\prime }={\displaystyle \frac{dy}{dx}={\displaystyle \frac{d(f(x))}{dx}=f^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}}}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}f(x)=2x,\text{hitunglah laju }\\ & \text{perubahan fungsi}f\text{di}x=2\\ & \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}f(x)=2x\\ & \begin{array}{|c|}\hline \text{Cara Pertama}\\ \begin{array}{rl}f(x)& =2x\\ f(2)& =2.2=4\\ f^{\prime }(2)& =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{f(x)-f(2)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{(2x)-(4)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{2x-4}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle 2}\\ & =2\end{array}\\ \text{Cara Kedua}\\ \begin{array}{rl}f(2)& =4\\ f(2+& h)=2(2+h)=4+2h\\ f(2+& h)-f(2)=2h\\ f^{\prime }(2)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\\ & =2\end{array}\\ \hline\end{array}\\ & \text{Jadi, laju perubahan fungsi}f\text{di}\\ & x=2\text{adalah}2\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika}f(x)=3x-5,\text{hitunglah laju }\\ & \text{perubahan fungsi}f\text{di}x=2\\ & \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}f(x)=3x-5\\ & \begin{array}{|c|}\hline \text{Cara Pertama}\\ \begin{array}{rl}f(x)& =3x-5\\ f(2)& =3.2-5=1\\ f^{\prime }(2)& =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{f(x)-f(2)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{(3x-5)-(1)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{3x-6}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle 3}\\ & =3\end{array}\\ \text{Cara Kedua}\\ \begin{array}{rl}f(2)& =1\\ f(2+& h)=3(2+h)-5=3h+1\\ f(2+& h)-f(2)=3h\\ f^{\prime }(2)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{3h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}3\\ & =3\end{array}\\ \hline\end{array}\\ & \text{Jadi, laju perubahan fungsi}f\text{di}\\ & x=2\text{adalah}3\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui}f(x)=2022x^2,\text{tentukanlah}{f}^{\prime }(x)\text{dan}{f}^{\prime }(1)\\ & \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline f^{\prime }(x)& f^{\prime }(1)\\ \begin{array}{rl}f(x)& =2022x^2\\ f(x+h)& =2022(x+h{)}^2\\ & =2022(x^2+2xh+h^2)\\ & =2022x^2+4044xh+2022h^2\\ f^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(2022x^2+4044xh+2022h^2)-\left(2022x^2\right)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4044xh+2022h^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 4044x+2022h}\\ & =4044x\end{array}& \begin{array}{rl}{f}^{\prime }(x)& =4044x\\ \text{maka},& \\ f^{\prime }(1)& =4044.1\\ & =4044\\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa fungsi}f(x)={\displaystyle \frac{1}{x^2}\text{dengan daerah asal}}\\ & {\text{D}}_f=\{x|x\in \mathbb{R},\text{dan}x\ne 0\}.\\ & \text{a}.\text{Tunjukkan bahwa}{f}^{\prime }(a)={\displaystyle -\frac{2}{a^3}}\\ & \text{b}.\text{jelaskanlah mengapa}{f}^{\prime }(0)\text{tidak terdefinisi}\\ & \\ & \text{Jawab}:\\ \\ & \begin{array}{|ll|}\hline \begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{1}{(x+h{)}^2}-{\displaystyle \frac{1}{x^2}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{x^2-(x+h{)}^2}{{(x(x+h))}^2}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{x^2-(x^2+2xh+h^2)}{h{(x(x+h))}^2}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2xh-h^2}{h{(x(x+h))}^2}}\\ & =\underset{h\to 0}{\text{Lim}}-{\displaystyle \frac{2x+h}{{(x(x+h))}^2}}\\ & =-{\displaystyle \frac{2x}{x^4}}\\ & =-{\displaystyle \frac{2}{x^3}}\end{array}& \begin{array}{rl}& \text{Untuk}\text{jawaban poin a dan b }\\ & \text{adalah sebagai berikut}\\ & f^{\prime }(x)=-{\displaystyle \frac{2}{x^3}}\\ & f^{\prime }(a)=-{\displaystyle \frac{2}{a^3}}\\ & \text{maka},\\ & f^{\prime }(0)=-{\displaystyle \frac{2}{0^3}}\\ & =-{\displaystyle \frac{2}{0}}\\ & \\ & \text{karena penyebut berupa }\\ & \text{bilangan}0\\ & \text{maka}{f}^{\prime }(0)\mathbf{\text{tidak terdefinisi}}\\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

Turunan Fungsi Aljabar

 A. Turunan Fungsi Aljabar

A. 1 Laju Perubahasan untuk Nilai Fungsi

Konsep turunan fungsi pada awalnya digunakan dalam bidang kususnya Matematika dan fisika, dalam hal hal ini kita berikan contohnya adalah laju perubahan kecepatan.

Coba perhatikanlah, misal pada kasus gerak jatuh bebas suatu benda yang dinyaatakan dengan  $h={\displaystyle \frac{1}{2}g{t}^2}$  dengan  $h$  adalah tinggi benda dengan percepatan grafitasinya adalah  $g=10m/s^2$ dan  $t$  adalah waktu tempuh.

Misalkan suatu benda jatuh dari ketinggian 125 meter dari permukaan tanah dengan percepatan grafitasinya adalah $g=10m/s^2$, maka waktu yang dibutuhkan benda tersebut untuk sampai ke tanah adalah:

$\begin{array}{rl}h& ={\displaystyle \frac{1}{2}g{t}^2}\\ 125& =\frac{1}{2}(10)t^2\\ 25& =t^2\\ 5& =t\end{array}$

Dari kejadian di atas dapat kita dapatkan kecepatan rata-ratanya yaitu: perubahan tinggi dibagi perubahan waktu terjadinya, atau misal dituliskan

$△v={\displaystyle \frac{△y}{△t}={\displaystyle \frac{y_n-y_1}{t_n-t_1}}}$

Sehingga kecepatan rata-ratanya adalah :  ${\displaystyle \frac{125}{5}=25m/s^2}$

Misalkan $f(t)$ untuk fungsi yang menujukkan posisi benda yang terjatuh dalam  $t$ dengan $f(t)=5t^2$, maka kecepatan rata-ratanya kita dapat menghitungnya untuk beberapa selang termasuk kita dapat menghitung kecepatan sesaatnya.

Coba perhatikanlah tabel berikut:

$\begin{array}{|ll|}\hline \{\begin{array}{ll}f(4)& ={5.4}^2=80\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{80-45}{4-3}}\\ & ={\displaystyle \frac{35}{1}=35}\end{array}\\ \{\begin{array}{ll}f(3,5)& =5.(3,5{)}^2=61,25\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{61,25-45}{3,5-3}}\\ & ={\displaystyle \frac{16,25}{0,5}=32,5}\end{array}\\ \{\begin{array}{ll}f(3,25)& =5.(3,25{)}^2=\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{52,8125-45}{3,25-3}}\\ & ={\displaystyle \frac{7,8125}{0,25}=31,25}\end{array}\\ \{\begin{array}{ll}f(3,1)& =5.(3,1{)}^2=48,05\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{48,05-45}{3,1-3}}\\ & ={\displaystyle \frac{3,05}{0,1}=30,5}\end{array}\\ \{\begin{array}{ll}f(3,1)& =5.(3,01{)}^2=45,3005\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{45,3005-45}{3,01-3}}\\ & ={\displaystyle \frac{0,3005}{0,01}=30,05}\end{array}\\ \hline\end{array}$

Dari ilsutrasi tabel di atas jika selisih waktu diperkecil terus menerus sampai mendekati nol, maka kecepatan sesaatnya akan mendekati nilai 30.

Sehingga kecepatan ketika $t=3$ ditentukan sebagai laju perubahan jarak terhadap waktu yang dibutuhkan dapat dituliskan dengan:

$\begin{array}{|cc|}\hline \text{Laju perubahan rata-rata}& \text{Laju perubahan sesaat}\\ \begin{array}{rl}& \\ & {\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}={\displaystyle \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}}}\\ & \end{array}& \begin{array}{rl}& \\ & \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(a+h)-f(a)}{h}}\\ & \end{array}\\ \hline\end{array}$.

Selanjutnya jika benda jatuh yang memenuhi kasus di atas, jika dihitung dengan pendekatan ini saat  $t=3$  adalah:

$\begin{array}{rl}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5(t+h{)}^2-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5(t^2+2th+h^2)-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5t^2+10th+5h^2-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{10th+5h^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}10t+5h\\ & =10t\end{array}$

Dari saat  $t=3$  kecepatan sesaatnya adalah $10t=10(3)=30m/s^2$.

Secara matematis, perubahan laju terhadap suatu fungsi di  $x=a$ selanjutnya dinotasikan dengan $f^{\prime }(x)$ dan didefiniskan dengan:

$\boxed{{\displaystyle f^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}}}$

Bentuk di atas dinamakan dengan derivatif atau turunan pertama pada fungsi  $f(x)$  dan dinotasikan dengan  $f^{\prime }(x)$ dan proses pencarian derivatif ini dinamakan differensial.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}g(x)=3x-5,\text{hitunglah laju perubahan }\\ & \text{fungsi}g\text{di}x=2\\ & \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}g(x)=3x-5\\ & \begin{array}{|cc|}\hline \text{Cara Pertama}& \text{Cara Kedua}\\ \begin{array}{rl}g(x)& =3x-5\\ g(2)& =3.2-5=1\\ g^{\prime }(2)& =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{g(x)-g(2)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{(3x-5)-(1)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{3x-6}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle 3}\\ & =3\end{array}& \begin{array}{rl}g(2)& =1\\ g(2+& h)=3(2+h)-5=3h+1\\ g(2+& h)-g(2)=3h\\ g^{\prime }(2)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{g(2+h)-g(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{3h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}3\\ & =3\\ & \end{array}\\ \hline\end{array}\\ & \text{Jadi, laju perubahan fungsi}g\text{di}x=2\text{adalah}3\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui}f(x)=2022x^2,\text{tentukanlah}{f}^{\prime }(x)\text{dan}{f}^{\prime }(1)\\ & \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline f^{\prime }(x)& f^{\prime }(1)\\ \begin{array}{rl}f(x)& =2022x^2\\ f(x+h)& =2022(x+h{)}^2\\ & =2022(x^2+2xh+h^2)\\ & =2022x^2+4044xh+2022h^2\\ f^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(2022x^2+4044xh+2022h^2)-\left(2022x^2\right)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4044xh+2022h^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 4044x+2022h}\\ & =4044x\end{array}& \begin{array}{rl}{f}^{\prime }(x)& =4044x\\ \text{maka},& \\ f^{\prime }(1)& =4044.1\\ & =4044\\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah kecepatan jika diketahui}f(t)=\sin t\\ & \text{saat}t\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{f}^{\prime }(t)=v(t)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (t+h)-\sin t}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}(2t+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\cos {\displaystyle \frac{1}{2}(2t+h).{\displaystyle \frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 2\cos {\displaystyle \frac{1}{2}(2t+h)\times {\displaystyle \frac{1}{2}}}}\\ & =2\cos {\displaystyle \frac{1}{2}(2t+0)\times {\displaystyle \frac{1}{2}}}\\ & =\cos {\displaystyle \frac{1}{2}(2t)}\\ & =\cos t\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Diketahui sebuah bola bergerak melingkar beraturan}\\ & \text{dengan persamaan}f(t)=2\sin 2t.\text{Tentukanlah}\\ & \text{kecepatan bola saat}t={\displaystyle \frac{1}{12}\pi }\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}v(t)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4\sin 2(t+h)-2\sin 2t}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4\cos {\displaystyle \frac{1}{2}(4t+2h)\sin {\displaystyle \frac{1}{2}(2h)}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}4\cos {\displaystyle \frac{1}{2}(4t+2h)\times \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h}}}\\ & =4\cos {\displaystyle \frac{1}{2}(4t)}\\ & =4\cos 2t\\ v\left({\displaystyle \frac{1}{12}\pi }\right)& =4\cos 2\left({\displaystyle \frac{1}{12}\pi }\right)\\ & =4\cos {\displaystyle \frac{1}{6}\pi }\\ & =4\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & =2\sqrt{3}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Noormandiri, B. K. 2004. Matematika SMA Jilid $2^A$ Berdasarkan Kurikulum 2004. Jakarta: ERLANGGA.
2. Noormandiri, B. K. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh 4 Soal dan Pembahasan Materi Lingkaran dan Hubungan Dua Lingkaran

 $\begin{array}{l}\\ 16.& \text{Salah satu garis singgung yang bersudut}{120}^{\circ }\\ & \text{terhadap sumbu x positif terhadap lingkaran}\\ & \text{dengan ujung diameter titik}(7,6)\text{dan}(1,-2)\\ & \text{adalah}....\\ & \text{a}.y=-x\sqrt{3}+4\sqrt{3}+12\\ & \text{b}.y=-x\sqrt{3}-4\sqrt{3}+8\\ & \text{c}.y=-x\sqrt{3}+4\sqrt{3}-4\\ & \text{d}.y=-x\sqrt{3}-4\sqrt{3}-8\\ & \text{e}.y=-x\sqrt{3}+4\sqrt{3}+22\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline \text{Pusat Lingkaran}& \text{Gradien Garis Singgung}\\ \begin{array}{rl}& (a,b)\\ & =\left({\displaystyle \frac{x_1+x_2}{2},\frac{y_1+y_2}{2}}\right)\\ & =\left({\displaystyle \frac{7+1}{2},\frac{6+(-2)}{2}}\right)\\ & =(4,2)\end{array}& \begin{array}{rl}m& =\tan {120}^{\circ }\\ & =-\tan ({180}^{\circ }-{60}^{\circ })\\ & =-\tan {60}^{\circ }\\ & =-\sqrt{3}\\ & \end{array}\\ \text{Jari-jari}& \text{Garis Singgung}\\ \begin{array}{rl}r& =\text{jarak titik}\\ & \text{singgung ke pusat}\\ & =\sqrt{(7-4{)}^2+(6-2{)}^2}\\ & =\sqrt{3^2+4^2}\\ & =\sqrt{25}\\ & =5\\ & \\ & \\ & \end{array}& \begin{array}{rl}& (y-b)=m(x-a)\pm r\sqrt{1+m^2}\\ & \iff (y-2)=-\sqrt{3}(x-4)\pm 5\sqrt{1+(-\sqrt{3}{)}^2}\\ & \iff y-2=-\sqrt{3}x+4\sqrt{3}\pm 5\sqrt{1+4}\\ & \iff y=-\sqrt{3}x+4\sqrt{3}+2\pm 10\\ & \iff y=\{\begin{array}{l}-\sqrt{3}x+4\sqrt{3}+2+10\\ -\sqrt{3}x+4\sqrt{3}+2-10\end{array}\\ & \iff y=\{\begin{array}{ll}-\sqrt{3}x+4\sqrt{3}+12& \\ -\sqrt{3}x+4\sqrt{3}-8& \end{array}\end{array}\\ \hline\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}$.

Dengan ilustrasi tambahan

$\begin{array}{l}\\ 17.& \text{Salah satu garis singgung lingkaran}\\ & x^2+y^2=10\text{yang ditarik dari}\\ & \text{titik}(4,2)\text{adalah}....\\ & \text{a}.x+3y=10\\ & \text{b}.x-3y=10\\ & \text{c}.-x-3y=10\\ & \text{d}.2x+y=10\\ & \text{e}.x+2y=10\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline \begin{array}{rl}& \text{Garis Singgung}\\ & \text{di titik}\\ & (x_1,y_1)=(4,2)\end{array}& \begin{array}{rl}& \text{Tahapan menentukan}\\ & \text{harga}m\\ & \end{array}\\ \begin{array}{rl}& y-y_1=m(x-x_1)\\ & y-2=m(x-4)\\ & y=mx-4m+2\\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}& x^2+y^2=10\\ & x^2+{(mx-4m+2)}^2=10\\ & x^2+m^2{x}^2+16m^2+4-8m^{2}x+4mx-16m=10\\ & x^2+m^2{x}^2+16m^2-8m^{2}x+4mx-16m-6=0\\ & (1+m^2)x^2+(4m-8m^2)x+16m^2-16m-6=0\\ & \{\begin{array}{ll}a& =1+m^2\\ b& =4m-8m^2\\ c& =16m^2-16m-6\end{array}\end{array}\\ \hline\end{array}\\ & \begin{array}{rl}& \text{Syarat menyinggung}D=0\\ & b^2-4ac=0\\ & {(4m-8m^2)}^2-4(1+m^2)(16m^2-16m-6)=0\\ & 16m^2-64m^3+64m^4-64m^2+64m+24-64m^4+64m^3+24m^2=0\\ & -24m^2+64m+24=0\\ & -3m^2+8m+3=0\\ & (m-3)(3m+1)=0\\ & m=3\text{atau}m=-{\displaystyle \frac{1}{3}}\\ & m=\{\begin{array}{ll}3& \Rightarrow y=3x-10\\ & \Rightarrow 3x-y=10\\ -{\displaystyle \frac{1}{3}}& \Rightarrow y=-{\displaystyle \frac{1}{3}x+\frac{4}{3}+2}\\ & \Rightarrow x+3y=10\end{array}\end{array}\end{array}$.

$.\begin{array}{rl}& \text{Berikut ilustrasi gambarnya}\end{array}$

$\begin{array}{l}\\ 18.& \text{Diketahui persamaan lingkaran}{x}^2+y^2=r^2\\ & \text{dan sebuah titik di luar lingkaran}M(a,b)\\ & \text{Posisi garis}ax+by=r^2\text{adalah}....\\ & \text{a}.\text{menyinggung lingkaran}\\ & \text{b}.\text{memotong lingkaran di dua titik}\\ & \text{c}.\text{melalui titik pusat lingkaran}\\ & \text{d}.\text{tidak memotong lingkaran}\\ & \text{e}.\text{tidak ada yang benar}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \bullet L\equiv x^2+y^2=r^2\\ & \bullet M(a,b)\text{di luar lingkaran}L\\ & \text{Selanjutnya perhatikan penjelasan berikut}\\ & \begin{array}{rl}& \text{Karena}M(a,b)\text{di luar lingkaran}L,\text{maka}\\ & \text{maka salah satu dari}a\text{atau}b\text{atau keduanya}\\ & \text{akan lebih besar nilanya dari pada}r.\\ & \text{Misalkan kita pilih}a>r\\ & \text{Ambil posisi saat memotong sumbu}-X,y=0\\ & \begin{array}{rl}& \text{Untuk lingkaran}{x}^2+y^2=r^2\\ & \bullet y=0\Rightarrow x^2+0^2=r^2\Rightarrow x=\left|r\right|\\ & \text{Untuk garis}ax+by=r^2\\ & \bullet y=0\Rightarrow ax=r^2\Rightarrow x={\displaystyle \frac{r^2}{a}}\\ & \text{Dari sini tampak posisi}x=\left|r\right|>{\displaystyle \frac{r^2}{a}\ge 0}\end{array}\\ & \text{Sehingga kesimpulannya adalah}:\\ & \text{garis tersebut akan selalu memotong lingkaran}\end{array}\\ & \mathbf{\text{Sebagai ilustrasi perhatikan gambar berikut}}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Dua lingkaran dengan persamaan}\\ & \text{lingkaran-lingkaran}{x}^2+y^2+6x-8y+21=0\\ & \text{dan}{x}^2+y^2+10x-8y+25=0\text{adalah}....\\ & \text{a}.\text{berpotongan di luar titik}\\ & \text{b}.\text{tidak berpotongan atau bersinggungan}\\ & \text{c}.\text{bersinggungan luar}\\ & \text{d}.\text{bersinggungan dalam}\\ & \text{e}.\text{sepusat}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{|ll|}\hline \text{Lingakaran}& \text{Pusat/r}\\ L_1\equiv x^2+y^2+6x-8y+21=0& \{\begin{array}{ll}{P}_1& =(-3,4)\\ r_1& =2\end{array}\\ L_2\equiv x^2+y^2+10x-8y+25=0& \{\begin{array}{ll}{P}_2& =(-5,4)\\ r_2& =4\end{array}\\ \hline\end{array}\\ & \text{dan}\\ & \begin{array}{|cc|}\hline \text{Jarak kedua pusat}& \text{Jumlah/selisih jari-jari}\\ \begin{array}{rl}& \left(P_1{P}_2\right)\\ & =\sqrt{(-3+5{)}^2+(4-4{)}^2}\\ & =\sqrt{2^2+0^2}=\sqrt{4}=2\end{array}& \begin{array}{r}\{\begin{array}{ll}{r}_1+r_2& =2+4=6\\ |r_1-r_2|& =|2-4|=2\end{array}\end{array}\\ \hline\end{array}\\ & \text{Karena nilai}{P}_1{P}_2=|r_1-r_2|=2\\ & \text{hal ini menunjukkan keduanya bersinggungan}\\ & \text{di dalam}\\ & \mathbf{\text{Sebagai ilustrasi perhatikan gambar berikut}}\end{array}$ .

$\begin{array}{l}\\ 20.& \text{Dua lingkaran dengan persamaan}\\ & \text{lingkaran-lingkaran}{x}^2+y^2+2x-6y+9=0\\ & \text{dan}{x}^2+y^2+8x-6y+9=0\text{adalah}....\\ & \text{a}.\text{berpotongan}\\ & \text{b}.\text{bersinggungan di dalam}\\ & \text{c}.\text{bersinggungan luar}\\ & \text{d}.\text{tidak berpotongan}\\ & \text{e}.\text{sepusat}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{|ll|}\hline \text{Lingakaran}& \text{Pusat/r}\\ L_1\equiv x^2+y^2+2x-6y+9=0& \{\begin{array}{ll}{P}_1& =(-1,3)\\ r_1& =1\end{array}\\ L_2\equiv x^2+y^2+8x-6y+9=0& \{\begin{array}{ll}{P}_2& =(-4,3)\\ r_2& =4\end{array}\\ \hline\end{array}\\ & \text{dan}\\ & \begin{array}{|cc|}\hline \text{Jarak kedua pusat}& \text{Jumlah/selisih jari-jari}\\ \begin{array}{rl}& \left(P_1{P}_2\right)\\ & =\sqrt{(-1+4{)}^2+(3-3{)}^2}\\ & =\sqrt{3^2+0^2}=\sqrt{9}=3\end{array}& \begin{array}{r}\{\begin{array}{ll}{r}_1+r_2& =1+4=5\\ |r_1-r_2|& =|1-4|=3\end{array}\end{array}\\ \hline\end{array}\\ & \text{Karena nilai}{P}_1{P}_2=|r_1-r_2|=3\\ & \text{hal ini menunjukkan keduanya bersinggungan}\\ & \text{di dalam}\\ & \mathbf{\text{Sebagai ilustrasi perhatikan gambar berikut}}\end{array}$.

DAFTAR PUSTAKA

1. Budi, W. S. 2010. Bahan Ajar Persiapan Menuju Olimpiade Sain Nasional/Internasional Matematika 3. Jakarta: ZAMRUD KEMALA.
2. Kartini, Suprapto, Subandi, dan Setiadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
3. Kanginan M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
4. Noormandiri. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
5. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU
6. Sukino. 2017. Matematika Jilid 2 untuk Kelas SMA/MA Kelas XI Kelompok Peminatan dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh 3 Soal dan Pembahasan Materi Lingkaran

 $\begin{array}{l}\\ 11.& \text{Lingkaran}{x}^2+y^2+2ax+2by+c=0\\ & \text{menyinggung sumbu Y jika}c=....\\ & \text{A}.ab\\ & \text{B}.a{b}^2\\ & \text{C}.a^{2}b\\ & \text{D}.a^2\\ & \text{E}.b^2\\ \\ & \mathbf{\text{Jawab}}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}& x^2+y^2+2ax+2by+c=0\\ & x=0\Rightarrow 0^2+y^2+2a.0+2by+c=0\\ & y^2+2by+c=0\{\begin{array}{ll}a& =1\\ b& =2b\\ c& =c\end{array}\\ & \text{Syarat menyinggung}\text{adalah}:\\ & D=b^2-4ac=0\\ & \iff (2b{)}^2-4.1.c=0\\ & \iff 4c=4b^2\\ & \iff c=b^2\end{array}\\ \\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& x^2+y^2+2ax+2by+c=0\\ & \iff x^2+2ax+a^2+y^2+2by+b^2+c-a^2-b^2=0\\ & \iff (x+a{)}^2+(y+b{)}^2=a^2+b^2-c\\ & \text{Karena menyinggung sumbu-Y, maka}R=a\\ & \text{Sehingga}{R}^2=a^2+b^2-c=a^2\\ & \iff b^2-c=0\\ & \iff b^2=c\\ & \iff c=b^2\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Diketahui pusat lingkaran L terletak dikuadran}\\ & \text{I dan berada di sepanjang garis}y=2x.\text{Jika}\\ & \text{lingkaran L menyinggung sumbu Y di titik}\\ & (0,6),\text{maka persamaan lingkaran L adalah}....\\ & \text{A}.x^2+y^2-3x-6y=0\\ & \text{B}.x^2+y^2+6x+12y-108=0\\ & \text{C}.x^2+y^2+12x+6y-72=0\\ & \text{D}.x^2+y^2-12x-6y=0\\ & \text{E}.x^2+y^2-6x-12y+36=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& (x-a{)}^2+(y-b{)}^2=r^2,\\ & \text{menyinggung titik}(0,6)\\ & \text{berarti pusat lingkaran L juga terletak}\\ & \text{pada garis}y=6.\text{Hal ini menunjukkan bahwa }\\ & \text{pusat lingkaran}\text{L berpusat di}(x,2x)=(\frac{y}{2},y),\\ & \text{dengan}y=6.\text{Dari informasi di atas, }\\ & \text{didapatlah pusat lingkaran berada di titik}(3,6).\\ & \text{Sehingga persamaan lingkarannya adalah}:\\ & (x-3{)}^2+(y-6{)}^2=3^2\text{ingat}r=\text{absis}x=3\\ & \iff (x-3{)}^2+(y-6{)}^2=x^2-6x+9+y^2+12x+36=9\\ & \iff x^2+y^2-6x+12y+36=0\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Persamaan garis singgung lingkaran}\\ & x^2+y^2+8x-3y-24=0,\text{di titik}\\ & (2,4)\text{adalah}....\\ & \text{A}.12x-5y-44=0\\ & \text{B}.12x+5y-44=0\\ & \text{C}.12x-y-50=0\\ & \text{D}.12x+y-50=0\\ & \text{E}.12x+y+50=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& x^2+y^2+8x-3y-24\\ & \iff x^2+8x+16+y^2-3y+{\displaystyle \frac{9}{4}-24=16+\frac{9}{4}}\\ & \iff (x+4{)}^2+(y-\frac{3}{2}{)}^2=16+\frac{9}{4}+24=42\frac{1}{4}\\ & \text{Persamaan garis singgung lingkar}\text{an lingkaran }\\ & \text{di titik}(x_1,y_1)\text{adalah}:\\ & (x_1+4)(x+4)+(y_1-\frac{3}{2})(y-\frac{3}{2})=42\frac{1}{4},\\ & \text{untuk}(x_1,y_1)=(2,4),\text{maka}\\ & (2+4)(x+4)+(4-\frac{3}{2})(y-\frac{3}{2})=\frac{169}{4}\\ & \iff 6(x+4)+\frac{5}{2}(y-\frac{3}{2})=\frac{169}{4}\\ & \iff 24(x+4)+5(2y-3)=169\\ & \iff 24x+96+10y-15=169\\ & \iff 24x+10y=169-96+15=88\\ & \iff 12x+5y-44=0\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Sebuah garis singgung}g\text{menyinggung }\\ & \text{lingkaran yang berpusat di}(-2,5)\text{dan}\\ & \text{berjari-jari}2\sqrt{10}\text{di titk}(4,3),\text{maka }\\ & \text{persamaan garis singgung}g\text{adalah}....\\ & \text{A}.y=3x+9\\ & \text{B}.y=3x-9\\ & \text{C}.y=-3x+9\\ & \text{D}.y=-3x-9\\ & \text{E}.y=3x+21\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& (x-a{)}^2+(y-b{)}^2=r^2\\ & \{\begin{array}{ll}\text{Pusat}& =(-2,5)\\ \text{r}& =2\sqrt{10}\end{array}\\ & \text{maka persamaan lingkarannya}:\\ & (x+2{)}^2+(y-5{)}^2=(2\sqrt{10}{)}^2\\ & \iff (x_1+2)(x+2)+(y_1-5)(y-5)=40,\\ & \text{menyingung garis}g\text{di}(4,3)\\ & (4+2)(x+2)+(3-5)(y-5)=40\\ & \iff 6x+12-2y+10=40\\ & \iff 6x-2y=40-12-10\\ & \iff 3x-y=9\\ & \iff -y=-3x+9\\ & \iff y=3x-9\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Suatu lingkaran dengan titik pusatnya terletak }\\ & \text{pada kurva}y=\sqrt{x}\text{dan melalui titik asal}O(0,0).\\ & \text{Jika diketahui absis titik pusat lingkaran tersebut }\\ & \text{adalah}a,\text{maka persamaan garis singgung }\\ & \text{lingkaran yang melalui titik}O\text{tersebut adalah}....\\ & \text{A}.y=-x\\ & \text{B}.y=-x\sqrt{a}\\ & \text{C}.y=-ax\\ & \text{D}.y=-2x\sqrt{2}\\ & \text{E}.y=-2ax\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|lcl|}\hline \begin{array}{rl}& \text{Pusat}\\ & \text{lingkaran}\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Gradien garis singgung}\\ & \text{yang tegak lurus dengan }\\ & \text{garis yang melalui titik}\\ & \text{pusat lingkaran yang }\\ & \text{bergradien}{m}_L\end{array}& \begin{array}{rl}& \text{Persamaan garis }\\ & \text{singgung yang }\\ & \text{melalui titik asal}\\ & O(0,0)\\ & \end{array}\\ \begin{array}{rl}& (a,b)\\ & =(a,\sqrt{a})\\ & \\ & \\ & \end{array}& \begin{array}{rl}& m.m_1=-1\\ & m.\frac{y}{x}=-1\\ & m=-\frac{x}{y}=-{\displaystyle \frac{a}{\sqrt{a}}}\\ & =-\sqrt{a}\end{array}& \begin{array}{rl}y& =mx,\\ & \text{karena melalui}\\ & \text{titik asal}\\ y& =-\sqrt{a}x,\\ y& =-x\sqrt{a}\end{array}\\ \hline\end{array}\end{array}$.

Contoh 2 Soal dan Pembahasan Materi Lingkaran

 $\begin{array}{l}\\ 6.& \text{Diketahui lingkaran}{x}^2+y^2+4x+ky-12=0\\ & \text{melalui titik}(-2,8)\text{maka jari-jari lingkaran}\\ & \text{tersebut adalah}....\\ & \text{A}.1\\ & \text{B}.5\\ & \text{C}.6\\ & \text{D}.12\\ & \text{E}.25\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui ingkaran berpusat di}(-2,-{\displaystyle \frac{1}{2}k}),\\ & \text{yaitu}:\\ & x^2+y^2+4x+ky-12=0\\ & \text{melalui}(-2,8)\text{berarti }\\ & (-2{)}^2+8^2+4(-2)+k.8-12=0\\ & 4+64-8-12+8k=0\\ & 48+8k=0\\ & k=-6\\ & \text{Sehingga}r=\sqrt{{\displaystyle \frac{4^2}{4}+\frac{(-6{)}^2}{4}-(-12)}}\\ & =\sqrt{{\displaystyle 4+9+12}}=\sqrt{25}=5\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Persmaan lingkaran}{x}^2+y^2+px+8y+9=0\\ & \text{menyinggung sumbu X. Pusat lingkaran tersebut }\\ & \text{adalah}....\\ & \text{A}.(6,-4)\\ & \text{B}.(6,6)\\ & \text{C}.(3,-4)\\ & \text{D}.(-6,-4)\\ & \text{E}.(3,4)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \mathbf{\text{Lingkaran}}{x}^2+y^2+px+8y+9=0\\ & \text{maka,}\\ & x^2+px+y^2+8y+9=0\\ & {(x+{\displaystyle \frac{1}{2}p})}^2-{\displaystyle \frac{1}{4}{p}^2+(y+4{)}^2-16+9=0}\\ & \iff {(x+{\displaystyle \frac{1}{2}p})}^2+(y+4{)}^2=7+{\displaystyle \frac{1}{4}{p}^2}\\ & \text{karena menyinggung sumbu-X,}R=b=4,\\ & \text{sehingga}\\ & 7+{\displaystyle \frac{1}{4}{p}^2=4^2}\\ & \iff {\displaystyle \frac{1}{4}{p}^2=16-7=9\iff p^2=36\iff p=\pm 6}\\ & p=-6\Rightarrow x^2+y^2-6x+8y+9=0\\ & \Rightarrow \text{pusatnya adalah}(-{\displaystyle \frac{A}{2},-\frac{B}{2}})=(3,-4)\\ & p=6\Rightarrow x^2+y^2+6x+8y+9=0\\ & \Rightarrow \text{pusatnya adalah}(-{\displaystyle \frac{A}{2},-\frac{B}{2}})=(-3,-4)\\ & \text{dan berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Titik-titik berikut yang posisinya berada di luar }\\ & \text{lingkaran}{x}^2+y^2-2x+8y-32=0\text{adalah}....\\ & \text{A}.(0,0)\\ & \text{B}.(-6,-4)\\ & \text{C}.(-3,2)\\ & \text{D}.(3,1)\\ & \text{E}.(4,1)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \begin{array}{|cclc|}\hline \text{Opsi}& \text{Titik}& \text{Lingkaran}& \text{Keterangan}\\ \text{A}& (0,0)& 0^2+0^2-2.0+8.0-32=-32& \text{dalam}\\ \text{B}& (-6,-4)& (-6{)}^2+(-4{)}^2-2(-6)+8(-4)-32=0& \text{pada}\\ \text{C}& (-3,2)& (-3{)}^2+(2{)}^2-2(-3)+8(2)-32=3& \mathbf{\text{di luar}}\\ \text{D}& (3,1)& 3^2+1^2-2.3+8.1-32=-20& \text{dalam}\\ \text{E}& (4,1)& 4^2+1^2-2.4+8.1-32=-15& \text{dalam}\\ \hline\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Diketahui garis}x-2y=5\text{memotong lingkaran}\\ & x^2+y^2-4y+8y+10=0\text{di titik A dan B}.\\ & \text{Panjang ruas garis AB adalah}....\\ & \text{A}.4\sqrt{2}\\ & \text{B}.2\sqrt{5}\\ & \text{C}.\sqrt{10}\\ & \text{D}.5\\ & \text{E}.4\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \begin{array}{rl}& \text{Perhatikanlah bahwa garis}x-2y=5\\ & \text{memotong lingkaran}\\ & x^2+y^2-4x+8y+10=0,\\ & \text{maka garis}x=2y+5\text{disubstitusikan ke}\\ & \text{lingkaran tersebut, yaitu}:\\ & (2y+5{)}^2+y^2-4(2y+5)+8y+10=0\\ & 4y^2+20y+25+y^2-8y-20+8y+10=0\\ & 5y^2+20y+15=0\\ & y^2+4y+3=0\\ & (y+1)(y+3)=0\\ & y=-1\vee y=-3\\ & \text{untuk nilai}\\ & y=-3\Rightarrow x=2(-3)+5=-1,A(-1,-3)\\ & y=-1\Rightarrow x=2(-1)+5=3,B(3,-1)\\ & \text{maka},\text{AB}=\sqrt{(3-(-1){)}^2+(-1-(-3){)}^2}\\ & =\sqrt{4^2+2^2}\\ & =\sqrt{16+4}\\ & =\sqrt{20}\\ & =2\sqrt{5}\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$ .

$\begin{array}{l}\\ 10.& \text{Kekhususan persamaan lingkaran}\\ & x^2+y^2-6x-6y+6=0\text{adalah}....\\ & \text{A}.\text{menyinggung sumbu X}\\ & \text{B}.\text{menyinggung sumbu Y}\\ & \text{C}.\text{berpusat di}O(0,0)\\ & \text{D}.\text{titik pusatnya terletak pada}x-y=0\\ & \text{E}.\text{berjari-jari 3}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui persamaan lingkaran}\\ & x^2+y^2-6x-6y+6=0\\ & x^2-6x+9+y^2-6y+9+6=9+9\\ & (x-3{)}^2+(y-3{)}^2=18-6\\ & (x-3{)}^2+(y-3{)}^2=12\\ & (x-3{)}^2+(y-3{)}^2={\left(2\sqrt{3}\right)}^2\\ & \text{lingkaran ini}\{\begin{array}{ll}\text{Pusat}& =(3,3)\\ \text{Jari-jari}& =2\sqrt{3}\end{array}\\ & \begin{array}{|clc|}\hline \text{Opsi}& \text{Pernyataan}& \text{Keterangan}\\ \text{A}& \text{menyinggung sumbu X}& \text{tidak tepat}\\ \text{B}& \text{menyinggung sumbu Y}& \text{tidak tepat}\\ \text{C}& \text{berpusat di}O(0,0)& \text{tidak tepat}\\ \text{D}& \text{titik pusatnya terletak pada garis}x-y=0& \mathbf{\text{tepat}}\\ \text{E}& \text{berjari-jari 3}& \text{tidak tepat}\\ \hline\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

Contoh 1 Soal dan Pembahasan Materi Lingkaran

 $\begin{array}{l}\\ 1.& \text{Jari-jari lingkaran dengan persamaan}{x}^2+y^2=48\\ & \text{adalah}....\\ & \text{A}.{\displaystyle 3\sqrt{5}}\\ & \text{B}.4\sqrt{3}\\ & \text{C}.5\sqrt{2}\\ & \text{D}.{\displaystyle 6\sqrt{3}}\\ & \text{E}.7\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{r}^2& =48\\ r& =\sqrt{48}\\ & =\sqrt{16.3}\\ & =4\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Titik pusat lingkaran}(x-7{)}^2+(y+9{)}^2=48\\ & \text{adalah}....\\ & \text{A}.{\displaystyle (-7,-9)}\\ & \text{B}.(-7,9)\\ & \text{C}.(7,-9)\\ & \text{D}.{\displaystyle (7,6)}\\ & \text{E}.(15,48)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Jelas bahwa}(a,b)& =(-6,9)\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Persamaan lingkaran yang berpusat di}P(-2,5)\\ & \text{dan melalui titik}T(3,4)\text{adalah}....\\ & \text{A}.(x+2{)}^2+(y-5{)}^2=26\\ & \text{B}.(x-3{)}^2+(y+5{)}^2=36\\ & \text{C}.(x+2{)}^2+(y-5{)}^2=82\\ & \text{D}.(x-3{)}^2+(y+5{)}^2=82\\ & \text{E}.(x+2{)}^2+(y+5{)}^2=82\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Persamaan Lingkaran Berpusat di}(a,b)\\ & \text{adalah}:(x-a{)}^2+(y-b{)}^2=r^2\\ & \begin{array}{|ll|}\hline \text{Pusat di}P(-2,5)& \text{Melalui Titik}T(3,4)\\ \begin{array}{rl}(x-a{)}^2+(y-b{)}^2& =r^2\\ (x+2{)}^2+(y-5{)}^2& =r^2\\ & \\ & \end{array}& \begin{array}{rl}(x-a{)}^2+(y-b{)}^2& =r^2\\ (3+2{)}^2+(4-5{)}^2& =r^2\\ 5^2+(-1{)}^2& =r^2\\ 26& =r^2\end{array}\\ \begin{array}{rl}& \text{Sehinga persamaan}\\ & \text{lingkarannya}\end{array}& \begin{array}{rl}& \text{adalah}:\\ & (x+2{)}^2+(y-5{)}^2=r^2=26\\ & (x+2{)}^2+(y-5{)}^2=26\\ & \end{array}\\ \hline\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Koordinat titik pusat dan jari-jari lingkaran}{x}^2+y^2-4x+6y+4=0\text{adalah}....\\ & \text{A}.(-3,2)\text{dan}3\\ & \text{B}.(3,-2)\text{dan}3\\ & \text{C}.(-2,-3)\text{ dan}3\\ & \text{D}.(2,-3)\text{dan}3\\ & \text{E}.(2,3)\text{dan}3\\ \\ & \mathbf{\text{Jawab}}:\\ & \mathbf{\text{Alterntif 1}}\\ & \begin{array}{|ll|}\hline & \text{Persamaan Lingkaran Berpusat di}(a,b)\text{dan berjari-jari}r\text{adalah}\\ & \begin{array}{rl}(x-a{)}^2+(y-b{)}^2& =r^2\\ x^2+y^2-4x+6y+4& =0\\ x^2-4x+y^2+6y+4& =0\\ x^2-4x+4-4+y^2+6y+9-9+4& =0\\ (x-2{)}^2-4+(y+3{)}^2-9+4& =0\\ (x-2{)}^2+(y+3{)}^2& =4+9-4\\ (x-2{)}^2+(y+3{)}^2& =9\\ (x-2{)}^2+(y-(-3){)}^2& =3^2\{\begin{array}{ll}\text{Pusat}& =(2,-3)\\ \text{dan}\\ r& =3\end{array}\end{array}\\ \hline\end{array}\\ & \mathbf{\text{Alterntif 2}}\\ & \begin{array}{rl}\text{Diketahui}& \text{persamaan lingkaran}:x^2+y^2-4x+6y+4=0\{\begin{array}{ll}A& =-4\\ B& =6\\ C& =4\end{array}\\ & x^2+y^2+Ax+By+C=0\\ & \{\begin{array}{ll}\text{Pusat}& =(-{\displaystyle \frac{1}{2}A,-\frac{1}{2}B})=(-\frac{1}{2}\cdots ,-\frac{1}{2}\cdots )=(\cdots ,\cdots )\\ \text{Jari-jari}& =\sqrt{{\displaystyle \frac{1}{4}{A}^2+\frac{1}{4}{B}^2-C}}=\sqrt{{\displaystyle \frac{1}{4}{\cdots }^2+\frac{1}{4}{\cdots }^2-\cdots }}=\sqrt{\cdots }\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Suatu lingkaran}{x}^2+y^2-4x+2y+p=0\\ & \text{berjari-jari 3, maka nilai}p\text{adalah}....\\ & \text{A}.-1\\ & \text{B}.-2\\ & \text{C}.-3\\ & \text{D}.-4\\ & \text{E}.-5\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}r=\sqrt{{\displaystyle \frac{A^2}{4}+\frac{B^2}{4}-C}}& =3\\ {\displaystyle \sqrt{\frac{(-4{)}^2}{4}+\frac{2^2}{4}-p}}& =3\\ {\displaystyle \frac{16}{4}+\frac{4}{4}-p}& =9\\ 4+1-p& =9\\ -p& =9-5\\ p& =-4\end{array}\end{array}$.

Hubungan Dua Lingkaran

Hubungan Dua Buah Lingkaran

Coba perhatikan ilustrasi beberapa lingkaran berikut

Sebagai penjelasan dari kondisi di atas adalah:

$\begin{array}{|ccl|}\hline \mathbf{\text{Kedudukan}}& \mathbf{\text{Ilustrasi}}& \mathbf{\text{Keterangan}}\\ \left|L_1{L}_2\right|>r_1+r_2& \mathbf{\text{Gambar 1}}& \begin{array}{rl}& \text{kedua lingkaran tidak berpotongan}\\ & \text{dan tidak pula bersinggungan}\\ & \text{dan saling lepas}\end{array}\\ \left|L_1{L}_2\right|=0& \mathbf{\text{Gambar 5}}& \text{Dikarenakan sepusat}\\ \left|L_1{L}_2\right|\le r_1+r_2& \mathbf{\text{Gambar 6}}& \text{Terletak di dalam lingkaran}{L}_1\\ \left|L_1{L}_2\right|=r_1+r_2& \mathbf{\text{Gambar 2}}& \begin{array}{rl}& \text{kedua lingkaran tidak berpotongan}\\ & \text{tetapi bersinggungan di luar}\end{array}\\ \left|L_1{L}_2\right|=r_1-r_2& \mathbf{\text{Gambar 3}}& \begin{array}{rl}& \text{kedua lingkaran tidak berpotongan}\\ & \text{tetapi bersinggungan di dalam}\end{array}\\ \{\begin{array}{l}\left|L_1{L}_2\right|>r_1-r_2\\ \left|L_1{L}_2\right|<r_1+r_2\end{array}& \mathbf{\text{Gambar 4}}& \begin{array}{rl}& \text{kedua lingkaran berpotongan}\end{array}\\ \hline\end{array}$.

$\begin{array}{rl}& \mathbf{\text{Kuasa}}\\ & \begin{array}{|ll|}\hline \text{Lingkaran}& \text{Posisi sebuah titik terhadap lingkaran}\\ \begin{array}{rl}& \text{Titik dua}\\ & \text{lingkaran}\end{array}& \begin{array}{rl}& \text{Tempat kedudukan titik-titik yang memiliki}\\ & \text{kuasa yang sama terhadap dua lingkaran}\end{array}\\ \begin{array}{rl}& \text{Garis tiga}\\ & \text{lingkaran}\end{array}& \begin{array}{rl}& \text{Tempat kedudukan titik yang memiliki}\\ & \text{kuasa yang sama terhadap tiga buah lingkaran}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{rl}& \mathbf{\text{Berkas Lingkaran}}\\ & \begin{array}{|lll|}\hline \text{Istilah}& \text{Posisi}& \text{Keterangan}\\ \begin{array}{rl}& \text{Berkas}\\ & \text{Lingkaran}\end{array}& \begin{array}{rl}& \text{Pada garis}\\ & \text{busur}\end{array}& \begin{array}{rl}& \text{Sejumlah lingkaran yang dapat }\\ & \text{dibuat melalui titik-titik potong }\\ & \text{kedua lingakaran itu}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{rl}& \mathbf{\text{Tali Busur Sekutu}}\\ & \begin{array}{|lll|}\hline \text{Istilah}& \text{Posisi}& \text{Keterangan}\\ \begin{array}{rl}& \text{Tali Busur}\\ & \text{Sekutu}\\ & \\ & \end{array}& \begin{array}{rl}& \text{Kedua}\\ & \text{lingkaran}\\ & \text{yang}\\ & \text{berpotongan}\end{array}& \begin{array}{rl}& \text{Ruas garis yang menghubungkan }\\ & \text{titik-titik potong irisan irisan }\\ & \text{kedua lingkaran tersebut}\\ & \end{array}\\ \hline\end{array}\\ & \bullet \text{Persamaan Tali Busur Sekutunya adalah}:L_1-L_2=0\\ & \bullet \text{Persamaan yang melalui titik potong dan lingkaran (berkas)}\\ & \text{itu adalah}:L_3=L_1+p(L_1-L_2),\text{atau}{L}_3=L_1+p{L}_2\\ & \text{dengan}p\text{adalah suatu parameter (suatu patokan nilai)}\\ & \bullet \text{Luas daerah irisan}:({\theta }_1{r}_1^2+{\theta }_2{r}_2^2)-{\displaystyle \frac{1}{2}(r_1^2\sin {\theta }_1+r_2^2\sin {\theta }_2)}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah kedudukan untuk dua buah lingkaran}\\ & L_1\equiv x^2+y^2-2x-4y+1=0\\ & \text{dan}{L}_2\equiv x^2+y^2-4x-2y-1=0.\\ & \text{Jika kedua lingkaran tersebut bersinggungan}\\ & \text{atau berpotongan, tentukanlah titik singgung atau potongnya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline L_1& L_2\\ x^2+y^2-2x-4y+1=0& x^2+y^2-4x-2y-1=0\\ \{\begin{array}{ll}{P}_1& :(-{\displaystyle \frac{1}{2}(-2),-\frac{1}{2}(-4)})=(1,2)\\ r& =\sqrt{{\displaystyle \frac{1}{4}((-2{)}^2+(-4^2))-1}}\\ & =2\end{array}& \{\begin{array}{ll}{P}_2& :(-{\displaystyle \frac{1}{2}(-4),-\frac{1}{2}(-2)})=(2,1)\\ r& =\sqrt{{\displaystyle \frac{1}{4}((-4{)}^2+(-2^2))-(-1)}}\\ & =\sqrt{6}\end{array}\\ \hline\end{array}\\ & \begin{array}{rl}\text{Jarak ke}& \text{dua pusat lingkarannya adalah}{P}_1{P}_2\text{yaitu}:\\ P_1{P}_2& =\sqrt{(2-1{)}^2+(1-2{)}^2}\\ & =\sqrt{2}\\ \text{Karena n}& \text{ilai}{P}_1{P}_2=\sqrt{2}\text{dan nilai}{P}_1+P_2=2+\sqrt{6},\\ \text{sehingga}& P_1{P}_2<P_1+P_2\text{maka kedua lingkaran }\\ \text{itu berpo}& \text{tongan}\end{array}\end{array}$.

$.\begin{array}{rl}{x}^2+y^2-2x-4y+1& =0..................(1)\\ x^2+y^2-4x-2y-1& =0..................(2)\\ ----------& ---{}^{-}\\ 2x-2y+2& =0\\ y& =x+1........................(3)\\ \text{persamaan}(3)\to (1)& \\ x^2+(x+1{)}^2-2x-& 4(x+1)+1=0\\ x^2+x^2+2x+1-2x& -4x-4+1=0\\ 2x^2-4x-2& =0\iff x^2-2x-1=0\\ x_{1,2}& ={\displaystyle \frac{-(-2)\pm \sqrt{(-2{)}^2-4.1(-1)}}{2}}\\ & ={\displaystyle \frac{2\pm \sqrt{8}}{2}={\displaystyle \frac{2\pm 2\sqrt{2}}{2}\{\begin{array}{ll}{x}_1& =1+\sqrt{2}.........(4)\mathbf{\text{atau}}\\ x_2& =1-\sqrt{2}.........(5)\end{array}}}\\ \text{persamaan}(4)\to (3)& ,y_1=1+\sqrt{2}+1=2+\sqrt{2}\\ \text{persamaan}(5)\to (3)& ,y_1=1-\sqrt{2}+1=2-\sqrt{2}\\ \text{Sehingga titik poton}& \text{gnya ada 2 yaitu}:\\ & \{\begin{array}{l}(1+\sqrt{2},2+\sqrt{2})\text{dan}\\ (1-\sqrt{2},2-\sqrt{2})\end{array}& \\ \text{Berikut ilustrasinya}\end{array}$.

$\begin{array}{l}\\ 2& \text{Dari contoh soal no.1, tentukanlah persamaan lingkaran }\\ & \text{yang melalui titik potong kedua lingkaran itu serta }\\ & \text{melalui titik pusat koordinat}O(0,0)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Pada jawaban soal no.1 didapatkan persamaan }\\ & \mathbf{\text{tali busur}}:L_1-L_2\equiv x-y+1=0\\ & \text{Sehingga persamaan }\\ & \mathbf{\text{berkas lingkaran}}\text{nya adalah}:L_3=L_1+p(L_1-L_2)=0\\ & \iff L_3=(x^2+y^2-2x-4y+1)+p(x-y+1)=0\\ & \text{Karena melalui titik asal}O(0,0),\text{maka}\\ & \iff (0+0-0-0+1)+p(0-0+1)=0\iff p=-1\\ & \text{Selanjutnya persamaan berkas lingkarannya akan menjadi}\\ & L_3\equiv x^2+y^2-2x-4y+1-(x-y+1)=0\\ & \text{Jadi},L_3\equiv x^2+y^2-3x-3y=0\\ \\ & \text{Dan gambar berikut sebagai ilustrasinya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3& \text{Diketahuin dua buah lingkaran}\\ & L_1\equiv x^2+y^2-15y+32=0\text{dan}\\ & L_2\equiv x^2+y^2-18x+2y+32=0\\ & \text{Tunjukkan bahwa kedua lingkaran}\\ & \text{bersinggungan di luar dan tentukan}\\ & \text{titik singgungnya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Akan ditunjukkan kedua lingkaran saling}\\ & \text{bersinggungan di luar, yaitu:}\\ & \begin{array}{|ll|}\hline \text{Lingakaran}& \text{Pusat/r}\\ L_1\equiv x^2+y^2-15y+32=0& \{\begin{array}{ll}{P}_1& =(0,8)\\ r_1& =4\sqrt{2}\end{array}\\ L_2\equiv x^2+y^2-18x+2y+32=0& \{\begin{array}{ll}{P}_2& =(9,-1)\\ r_2& =5\sqrt{2}\end{array}\\ \hline\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{|ll|}\hline \text{Hitungan jarak kedua pusat}& \text{Sebagai bandingan}\\ \begin{array}{rl}& \text{Pusat 1 lingkaran}{P}_1=(0,8)\\ & \text{Pusat 2 lingkaran}{P}_2=(9,-1)\\ & \text{maka jarak}{P}_1{P}_2\text{adalah}\\ & =\sqrt{(9-0{)}^2+(-1-8{)}^2}\\ & =\sqrt{9^2+9^2}=\sqrt{2\times 9^2}=9\sqrt{2}\end{array}& \begin{array}{rl}{P}_1{P}_2& =r_1+r_2\\ & =4\sqrt{2}+5\sqrt{2}\\ & =9\sqrt{2}\\ & \\ & \end{array}\\ \hline\end{array}\\ & \text{Adapun koordinat titik singgungnya}:\\ & \begin{array}{rl}\left(\begin{array}{c}x\\ y\end{array}\right)& ={\displaystyle \frac{5\left(\begin{array}{c}0\\ 8\end{array}\right)+4\left(\begin{array}{c}9\\ -1\end{array}\right)}{5+4}={\displaystyle \frac{\left(\begin{array}{c}5\times 0+4\times 9\\ 5\times 8+4\times (-1)\end{array}\right)}{9}}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}36\\ 36\end{array}\right)}{9}=\left(\begin{array}{c}4\\ 4\end{array}\right)}\end{array}\\ & \text{Sehingga koordinat titik potongnya adalah}:(4,4)\\ & \mathbf{\text{Sebagai gambaran perhatikan ilustrasi berikut}}\end{array}$.

DAFTAR PUSTAKA

1. Kanginan M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Noormandiri. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
3. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU
4. Sukino. 2017. Matematika Jilid 2 untuk Kelas SMA/MA Kelas XI Kelompok Peminatan dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh 7 Soal dan Pembahasan Materi Vektor

$\begin{array}{l}\\ 31.& \text{Diketahui}\left|\overrightarrow{a}\right|=\sqrt{3},\left|\overrightarrow{b}\right|=1,\text{dan}|\overrightarrow{a}-\overrightarrow{b}|=1\\ & \text{maka panjang vektor}\overrightarrow{a}+\overrightarrow{b}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \sqrt{3}& & & \text{d}.& 2\sqrt{2}\\ \text{b}.& \sqrt{5}& \text{c}.& \sqrt{7}& \text{e}.& 3\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Diketahui}& \text{sebagaimana pada soal}\\ {|\overrightarrow{a}-\overrightarrow{b}|}^2& ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ 1^2& ={\left(\sqrt{3}\right)}^2+1^2-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta & =3\\ \text{maka pan}& \text{jang vektor}\overrightarrow{a}+\overrightarrow{b}\text{adalah}\\ |\overrightarrow{a}+\overrightarrow{b}|& =\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta }\\ & =\sqrt{{\left(\sqrt{3}\right)}^2+1^2+3}\\ & =\sqrt{3+1+3}\\ & =\sqrt{7}\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Jika}\left|\bar{u}\right|=6,\left|\bar{v}\right|=4\sqrt{3},\text{dan}|\bar{u}-\bar{v}|=8\\ & \text{tentukanlah nilai dari}\\ & \text{a}.\bar{u}\bullet \bar{v}\\ & \text{b}.|\bar{u}+\bar{v}|\\ & \text{c}.\mathbf{\text{cosinus}}\text{sudut antara}\bar{u}\text{dan}\bar{v}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \bar{u}\bullet \bar{v}=\cdots \\ & 2.\bar{u}\bullet \bar{v}={\left|\bar{u}\right|}^2+{\left|\bar{v}\right|}^2-{|\bar{u}-\bar{v}|}^2\\ & 2.\bar{u}\bullet \bar{v}=6^2+(4\sqrt{3}{)}^2-8^2\\ & 2.\bar{u}\bullet \bar{v}=36+48-64=84-64=20\\ & \bar{u}\bullet \bar{v}={\displaystyle \frac{20}{2}=10}\\ \text{b}.& {|\bar{u}+\bar{v}|}^2={\left|\bar{u}\right|}^2+{\left|\bar{v}\right|}^2+2.\bar{u}\bullet \bar{v}\\ & {|\bar{u}+\bar{v}|}^2=6^2+(4\sqrt{3}{)}^2+20\\ & =84+20=104\\ & |\bar{u}+\bar{v}|=\sqrt{104}\\ \text{c}.& \cos \mathrm{\angle }(\bar{u},\bar{v})={\displaystyle \frac{\bar{u}\bullet \bar{v}}{\left|\bar{u}\right|.\left|\bar{v}\right|}=\frac{10}{6.(4\sqrt{3})}\times \frac{\sqrt{3}}{\sqrt{3}}}\\ & ={\displaystyle \frac{10\sqrt{3}}{72}=\frac{5}{36}\sqrt{3}}\\ & \mathbf{\text{Berikut ilustrasi gambarnya}}\end{array}\end{array}$.

$\begin{array}{l}\\ 33.& \text{Jika}\overrightarrow{p}=\left(\begin{array}{c}2\\ -5\end{array}\right),\overrightarrow{q}=\left(\begin{array}{c}4\\ 3\end{array}\right),\text{maka }\\ & \text{proyeksi skalar ortogonal vektor}\overrightarrow{p}\\ & \text{pada}\overrightarrow{q}\text{adalah}....\\ & \text{a}.{\displaystyle \frac{3}{5}}\\ \\ & \text{b}.{\displaystyle \frac{7}{5}}\\ \\ & \text{c}.{\displaystyle \frac{8}{5}}\\ \\ & \text{d}.{\displaystyle \frac{9}{5}}\\ \\ & \text{e}.{\displaystyle 2}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\left|\overrightarrow{r}\right|& ={\displaystyle \frac{\overrightarrow{p}.\overrightarrow{q}}{\left|\overrightarrow{q}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -5\end{array}\right).\left(\begin{array}{c}4\\ 3\end{array}\right)}{\sqrt{4^2+3^2}}}\\ & ={\displaystyle \frac{8+(-15)}{\sqrt{25}}}\\ & =\left|{\displaystyle \frac{-7}{5}}\right|\\ & ={\displaystyle \frac{7}{5}}\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Panjang Proyeksi vektor}\overrightarrow{a}=\left(\begin{array}{c}5\\ 1\end{array}\right)\\ & \text{pada}\overrightarrow{b}=\left(\begin{array}{c}0\\ 4\end{array}\right)\text{adalah}....\\ & \text{a}.{\displaystyle -1}\\ & \text{b}.-{\displaystyle \frac{1}{2}}\\ & \text{c}.1\\ & \text{d}.{\displaystyle 2}\\ & \text{e}.4\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\left|\overrightarrow{c}\right|& =\left|{\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right|\\ & =\left|{\displaystyle \frac{\left(\begin{array}{c}5\\ 1\end{array}\right)\bullet \left(\begin{array}{c}0\\ -4\end{array}\right)}{\left|\sqrt{0^2+(-4{)}^2}\right|}}\right|\\ & =\left|{\displaystyle \frac{0-4}{4}}\right|=|-1|=1\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Proyeksi vektor ortogonal}\overrightarrow{a}=\left(\begin{array}{c}2\\ -4\end{array}\right)\\ & \text{pada}\overrightarrow{b}=\left(\begin{array}{c}-1\\ 2\end{array}\right)\text{adalah}....\\ & \text{a}.\left(\begin{array}{c}2\\ -1\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}2\\ -2\end{array}\right)\\ & \text{c}.\left(\begin{array}{c}2\\ -4\end{array}\right)\\ & \text{d}.\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ & \text{e}.\left(\begin{array}{c}-2\\ 4\end{array}\right)\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{c}& =\left({\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}}\right).\overrightarrow{b}\\ & =\left({\displaystyle \frac{\left(\begin{array}{c}2\\ -4\end{array}\right)\bullet \left(\begin{array}{c}-1\\ 2\end{array}\right)}{(-1{)}^2+2^2}}\right).\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ & =\left({\displaystyle \frac{-2-8}{1+4}}\right).\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ & =-2\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}2\\ -4\end{array}\right)\end{array}\end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
2. Kusnandar, Muharman, I., Indrianti, M. 2017. Pendalaman Buku Teks Matematika SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.
3. Miyanto, Aksin, N., Suparno. 2021. Buku Interaktif Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas X Semester 2. Yogyakarta: INTAN PARIWARA. 
4. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
5. Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

Contoh 6 Soal dan Pembahasan Materi Vektor

 $\begin{array}{l}\\ 26.& \text{Jika}\overrightarrow{p}=\left(\begin{array}{c}-2\\ 4\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}8\\ 4\end{array}\right),\text{maka}\\ & \text{sudut yang dibentuk vektor}\overrightarrow{p}\text{dan}\overrightarrow{q}\\ & \text{adalah}....\\ & \text{a}.0^{\circ }\\ & \text{b}.{60}^{\circ }\\ & \text{c}.{45}^{\circ }\\ & \text{d}.{60}^{\circ }\\ & \text{e}.{90}^{\circ }\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}.\overrightarrow{q}& ={\displaystyle \left|\begin{array}{c}\overrightarrow{p}\end{array}\right|.\left|\begin{array}{c}\overrightarrow{q}\end{array}\right|.\cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})}\\ \cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& ={\displaystyle \frac{\overrightarrow{p}.\overrightarrow{q}}{\left|\overrightarrow{p}\right|.\left|\overrightarrow{q}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-2\\ 1\end{array}\right).\left(\begin{array}{c}8\\ 4\end{array}\right)}{\sqrt{(-2{)}^2+1^2}.\sqrt{8^2+4^2}}}\\ & ={\displaystyle \frac{-16+16}{\sqrt{20}.\sqrt{80}}}\\ & ={\displaystyle \frac{0}{40}}\\ & =0\\ \cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& =\cos {90}^{\circ }\\ \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& ={90}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Jika}\overline{OA}=\left(\begin{array}{c}1\\ 2\end{array}\right),\overline{OB}=\left(\begin{array}{c}4\\ 2\end{array}\right),\text{dan}\\ & \theta =\mathrm{\angle }(\overline{OA},\overline{OB}),\text{maka}\tan \theta =....\\ \\ & \text{a}.{\displaystyle \frac{3}{5}}\\ \\ & \text{b}.{\displaystyle \frac{9}{16}}\\ \\ & \text{c}.{\displaystyle \frac{3}{4}}\\ \\ & \text{d}.{\displaystyle \frac{4}{3}}\\ \\ & \text{e}.{\displaystyle \frac{16}{9}}\\ \\ & \text{Jawab}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 2\end{array}\right).\left(\begin{array}{c}4\\ 2\end{array}\right)}{\sqrt{1^2+2^2}\sqrt{4^2+2^2}}}\\ & ={\displaystyle \frac{4+4}{\sqrt{5}.\sqrt{20}}}\\ & ={\displaystyle \frac{8}{10}}\end{array}& \begin{array}{rl}\sin \theta & =\sqrt{1-{\cos }^2\theta }\\ & =\sqrt{1-{\left({\displaystyle \frac{8}{10}}\right)}^2}\\ & =\sqrt{{\displaystyle \frac{36}{100}}}\\ & ={\displaystyle \frac{6}{10}}\\ & \end{array}\\ \hline\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}\tan \theta & ={\displaystyle \frac{\sin \theta }{\cos \theta }}\\ & ={\displaystyle \frac{{\displaystyle \frac{6}{10}}}{{\displaystyle \frac{8}{10}}}}\\ & ={\displaystyle \frac{3}{4}}\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Jika}\overrightarrow{a},\overrightarrow{b}\text{dan}\overrightarrow{c}\text{adalah vektor satuan dengan}\\ & \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\text{Nilai dari}\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}\text{adalah}....\\ & \text{a}.{\displaystyle -3}\\ & \text{b}.{\displaystyle -\frac{3}{2}}\\ & \text{c}.{\displaystyle 0}\\ & \text{d}.{\displaystyle \frac{3}{2}}\\ & \text{e}.{\displaystyle 3}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Karena}& \{\begin{array}{c}\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{adalah vektor satuan, dan}\\ \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\end{array}\\ \text{segitig}& \text{a ABC adalah segitiga sama sisi}\\ \overrightarrow{a}.\overrightarrow{b}=& \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \overrightarrow{a}.\overrightarrow{c}=& \left|\overrightarrow{a}\right|\left|\overrightarrow{c}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \overrightarrow{b}.\overrightarrow{c}=& \left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \text{Jadi, ni}& \text{lai dari}\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}\\ & =(-\frac{1}{2})+(-\frac{1}{2})+(-\frac{1}{2})=-\frac{3}{2}\end{array}\end{array}$

$.\text{berikut ilustrasinya}$

$\begin{array}{l}\\ 29.& \text{Jika}\mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})={60}^{\circ },\left|\overrightarrow{a}\right|=4\text{dan}\\ & \left|\overrightarrow{b}\right|=3,\text{maka}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})\text{adalah}....\\ & \text{a}.2\\ & \text{b}.4\\ & \text{c}.6\\ & \text{d}.8\\ & \text{e}.10\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})& =\overrightarrow{a}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}\\ & =\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|\cos 0^{\circ }-\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos {60}^{\circ }\\ & ={\left|\overrightarrow{a}\right|}^2-\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|.{\displaystyle \frac{1}{2}}\\ & =4^2-4.3.{\displaystyle \frac{1}{2}}\\ & =16-6\\ & =10\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Tentukan}\bar{u}\bullet \bar{v},\text{jika diketahui}\\ & \text{a}.\left|\bar{u}\right|=10,\left|\bar{v}\right|=8\sqrt{3},\cos \mathrm{\angle }(\bar{v},\bar{u})={\displaystyle \frac{2}{5}\sqrt{3}}\\ & \text{b}.\left|\bar{u}\right|=6\sqrt{3},\left|\bar{v}\right|=4\sqrt{2},\cos (\bar{v},\bar{u})={30}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \bar{u}\bullet \bar{v}=\left|\bar{u}\right|.\left|\bar{v}\right|.\cos \mathrm{\angle }(\bar{v},\bar{u})\\ & \bar{u}\bullet \bar{v}=10.(8\sqrt{3}).{\displaystyle \frac{2}{5}\sqrt{3}=2.8.2.3=96}\\ \text{b}.& \bar{u}\bullet \bar{v}=\left|\bar{u}\right|.\left|\bar{v}\right|.\cos \mathrm{\angle }(\bar{v},\bar{u})\\ & \bar{u}\bullet \bar{v}=(6\sqrt{3}).(4\sqrt{2}).{\displaystyle \cos {30}^{\circ }}\\ & =(6\sqrt{3}).(4\sqrt{2}).{\displaystyle \frac{1}{2}\sqrt{3}}\\ & ={\displaystyle \frac{6.4.3.\sqrt{2}}{2}=36\sqrt{2}}\end{array}\end{array}$.