Contoh Soal 2 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 6.& \text{Jika}W=\sin 2t,\text{maka}{\displaystyle \frac{dW}{dt}=....}\\ & \begin{array}{l}\\ \text{a}.\cos 2t& \text{d}.2t\cos 2t+\sin 2t\\ \text{b}.2\cos 2t& \text{e}.\sin 2t-t\cos 2t\\ \text{c}.\sin 2t+t\cos 2t\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & W=\sin 2t\\ & W=\sin u\text{dengan}u=2t\\ & {\displaystyle \frac{dW}{dt}={\displaystyle \frac{dW}{du}.\frac{du}{dt}}}\\ & =\cos u.2\\ & =2\cos u\\ & =2\cos 2t\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}f(x)={\displaystyle \frac{2x+4}{1+\sqrt{x}},\text{maka}{\displaystyle f{}^{\prime }(4)=....}}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}}& & \text{d}.1\\ \text{b}.{\displaystyle \frac{3}{7}}& \text{c}.{\displaystyle \frac{3}{5}}& \text{e}.4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{2x+4}{1+\sqrt{x}}}\\ & ={\displaystyle \frac{U}{V}}\\ f{}^{\prime }(x)& ={\displaystyle \frac{U{}^{\prime }.V-U.V{}^{\prime }}{V^2}}\\ & ={\displaystyle \frac{(2)(1+\sqrt{x})-(2x+4).(1.{(1+\sqrt{x})}^0.{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}})}{{(1+\sqrt{x})}^2}}\\ & \text{ingat}\sqrt{x}=x^{{}^{\frac{1}{2}}}\\ f{}^{\prime }(4)& ={\displaystyle \frac{2(1+\sqrt{4})-(2.4+4).\frac{1}{2}.4^{{}^{-\frac{1}{2}}}}{{(1+\sqrt{4})}^2}}\\ & ={\displaystyle \frac{2(1+2)-(12).\frac{1}{2}.\frac{1}{2}}{(1+2{)}^2}}\\ & \text{ingat juga}{4}^{{}^{-\frac{1}{2}}}={\left(2^2\right)}^{{}^{-\frac{1}{2}}}=2^{{}^{-1}}={\displaystyle \frac{1}{2^1}=\frac{1}{2}}\\ & ={\displaystyle \frac{6-3}{9}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Jika}f(x)={\displaystyle \frac{1+\sin x}{\cos x},\text{maka}f{}^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}}& & \text{d}.2\\ \text{b}.{\displaystyle -\frac{1}{2}}& \text{c}.{\displaystyle \frac{3}{4}}& \text{e}.-2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{1+\sin x}{\cos x}}\\ f{}^{\prime }(x)& ={\displaystyle \frac{\cos x.\cos x-(1+\sin x).-\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}}\\ & \text{ingat bahwa}{\sin }^{2}x+{\cos }^{2}x=1\\ & ={\displaystyle \frac{1+\sin x}{{\cos }^{2}x}}\\ f{}^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)& ={\displaystyle \frac{1+\sin \left({\displaystyle \frac{1}{6}\pi }\right)}{{\cos }^2\left({\displaystyle \frac{1}{6}\pi }\right)}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{1}{2}}}{{\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}^2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{3}{4}}}}\\ & ={\displaystyle \frac{3}{2}\times \frac{4}{3}}\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}f(x)={\displaystyle 3x^2-2ax+7}\\ & \text{dan}f{}^{\prime }(1)=0,\text{maka}f{}^{\prime }(2)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 1}& & \text{d}.6\\ \text{b}.{\displaystyle 2}& \text{c}.{\displaystyle 4}& \text{e}.8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle 3x^2-2ax+7}\\ f{}^{\prime }(x)& =6x-2a\\ f{}^{\prime }(1)& =0\\ 6(1)-2a& =0\\ 6& =2a\\ 3& =a\\ \text{sehingga}& \\ f{}^{\prime }(x)& =6x-6\\ \text{maka},& \\ f{}^{\prime }(2)& =6.2-6\\ & =12-6\\ & =6\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jika}f(x)={\displaystyle (6x-3{)}^3(2x-1)}\\ & \text{maka}f{}^{\prime }(1)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 18}& & \text{d}.162\\ \\ \text{b}.{\displaystyle 24}& \text{c}.{\displaystyle 54}& \text{e}.216\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle (6x-3{)}^3(2x-1)}\\ & =(3.(2x-1){)}^3(2x-1{)}^1\\ & =3^3.(2x-1{)}^{3+1}\\ & =27(2x-1{)}^4\\ f{}^{\prime }(x)& =4.27(2x-1{)}^{4-1}.2\\ & =216.(2x-1{)}^3\\ f{}^{\prime }(1)& =216.(2.1-1{)}^3\\ & =216.1\\ & =216\end{array}\end{array}$.