Contoh Soal 5 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 21.& \text{Turunan pertama dari fungsi}\\ & g(x)={\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}\\ \text{b}.& {\displaystyle \frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}}\\ \text{c}.& {\displaystyle \frac{1}{{\sin }^{2}x{\cos }^{2}x}}\\ \text{d}.& {\displaystyle \frac{-1}{{\sin }^{2}x{\cos }^{2}x}}\\ \text{e}.& {\displaystyle {\sin }^{2}x{\cos }^{2}x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ g(x)& ={\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}\\ & =\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}={\displaystyle \frac{1}{\sin x\cos x}}\\ \text{maka}& \\ g^{\prime }(x)& ={\displaystyle \frac{0.(\sin x\cos x)-1.({\cos }^{2}x-{\sin }^{2}x)}{(\sin x\cos x{)}^2}}\\ & ={\displaystyle \frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Diketahui}h(x)=\cos \left({\displaystyle \frac{3}{x}}\right),\\ & \text{maka}{\displaystyle \frac{dh}{dx}}\\ & \begin{array}{l}\\ \text{a}.& -3\sin {\displaystyle \frac{3}{x}}\\ \text{b}.& -{\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\\ \text{c}.& -{\displaystyle \frac{3}{x}\sin \frac{3}{x}}\\ \text{d}.& {\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\\ \text{e}.& {\displaystyle \frac{3}{x}\sin \frac{3}{x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\cos {\displaystyle \frac{3}{x}}& =-\sin {\displaystyle \frac{3}{x}\left({\displaystyle \frac{0.(x)-3.1}{x^2}}\right)}\\ & ={\displaystyle \frac{-(-3)}{x^2}\sin \frac{3}{x}}\\ & ={\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Turunan pertama dari}\tan (\cos x),\\ & \text{terhadap}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\sec }^2(\cos x)\sin x\\ \text{b}.& {\sec }^2(\cos x)\sin x\\ \text{c}.& {\sec }^2(\sin x)\cos x\\ \text{d}.& {\displaystyle \sin x}\\ \text{e}.& {\displaystyle -\sin x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Misal}& \text{kan}\\ y& =\tan x(\cos x)\\ y^{\prime }& ={\sec }^2(\cos x)\times (-\sin x)\\ & =-{\sec }^2(\cos x).\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 24.& (\mathbf{\text{UN 2005}})\text{Turunan pertama dari}\\ & f(x)=\sqrt[3]{{\cos }^2(3x^2+5x)}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ \text{b}.& {\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)}\\ \text{c}.& -{\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ \text{d}.& -{\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\\ \text{e}.& {\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Misal}& \text{kan}\\ f(x)& =\sqrt[3]{{\cos }^2(3x^2+5x)}\\ f^{\prime }(x)& ={\cos }^{{.}^{\frac{2}{3}}}(3x^2+5x)\\ & ={\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{2}}}(3x^2+5x)\times (-\sin (3x^2+5x))}\\ & \times (6x+5)\\ & =-{\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ & =-{\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{\frac{2}{3}}}(3x^2+5x)}\\ & \times {\cos }^{-1}(3x^2+5x)\times \sin (3x^2+5x)\\ & =-{\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\end{array}\end{array}$