Sifat Turunan Pertama dan Aturan Rantai pada Turunan Fungsi Aljabar

Rumus Turunan dan Sifat Turunan Pertama

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

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

$\text{CONTOH SOAL}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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