Soal dan Pembahasan - Turunan Fungsi Trigonometri

Diperbarui 5 September 2021 — 18 Soal

Turunan fungsi trigonometri merupakan salah satu materi matematika yang dipelajari pada jenjang SMA, tepatnya di kelas XI. Berikut ini kami sajikan soal-soal yang berkaitan dengan materi turunan fungsi trigonometri, yang disertai dengan pembahasan.

Soal dan Pembahasan

✶ Nomor 1

Tentukan $D_xy$, jika diketahui $y=\sin x$.

Misalkan $f(x) = \sin x$, sehingga $$f(\textcolor{maroon}{x+h}) = \sin (\textcolor{maroon}{x+h})$$ Berdasarkan definisi turunan fungsi, diperoleh $$\begin{aligned} D_xy &= f'(x) \\ &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{\textcolor{green}{\sin (x+h)}-\textcolor{blue}{\sin x}}{h} \\ &= \lim_{h \to 0} \frac{(\sin x \cos h+\cos x \sin h)-\sin x}{h} \\ &= \lim_{h \to 0} \frac{\sin x \cos h-\sin x+\cos x \sin h}{h} \\ &= \lim_{h \to 0} \frac{\sin x (\cos h-1)+\cos x \sin h}{h} \\ &= \lim_{h \to 0} \frac{\sin x (\cos h-1)}{h}+\lim_{h \to 0} \frac{\cos x \sin h}{h} \\ &= \sin x \cdot \textcolor{red}{\lim_{h \to 0} \frac{\cos h-1}{h}}+\cos x \cdot \textcolor{red}{\lim_{h \to 0} \frac{\sin h}{h}} \end{aligned}$$

Karena $$\lim_{h \to 0} \frac{\cos h-1}{h}=0 \quad \text{dan} \quad \lim_{h \to 0} \frac{\sin h}{h}$$ maka $$\begin{aligned} f'(x) &= \sin x \cdot \textcolor{red}{0}+\cos x \cdot \textcolor{red}{1} \\ &= 0+\cos x \\ &= \cos x \end{aligned}$$

Pembahasan
✶ Nomor 2

Tentukan $D_xy$, jika diketahui $y=\cos x$.

Misalkan $f(x) = \cos x$, sehingga $$f(\textcolor{maroon}{x+h}) = \cos (\textcolor{maroon}{x+h})$$ Berdasarkan definisi turunan fungsi, diperoleh $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{\textcolor{green}{\cos (x+h)}-\textcolor{blue}{\cos x}}{h} \\ &= \lim_{h \to 0} \frac{(\cos x \cos h-\sin x \sin h)-\cos x}{h} \\ &= \lim_{h \to 0} \frac{\cos x \cos h-\cos x-\sin x \sin h}{h} \\ &= \lim_{h \to 0} \frac{\cos x (\cos h-1)-\sin x \sin h}{h} \\ &= \lim_{h \to 0} \frac{\cos x (\cos h-1)}{h}-\lim_{h \to 0} \frac{\sin x \sin h}{h} \\ &= \cos x \cdot \textcolor{red}{\lim_{h \to 0} \frac{\cos h-1}{h}}-\sin x \cdot \textcolor{red}{\lim_{h \to 0} \frac{\sin h}{h}} \end{aligned}$$

Karena $$\lim_{h \to 0} \frac{\cos h-1}{h}=0 \quad \text{dan} \quad \lim_{h \to 0} \frac{\sin h}{h}$$ maka $$\begin{aligned} f'(x) &= \cos x \cdot \textcolor{red}{0}-\sin x \cdot \textcolor{red}{1} \\ &= 0-\sin x \\ &= -\sin x \end{aligned}$$

Pembahasan
✶ Nomor 3

Tentukan hasil dari $D_x \tan x$.

Pertama, nyatakan $\tan x$ sebagai hasil bagi antara $\sin x$ dan $\cos x$. $$D_x \tan x = D_x \left( \frac{\sin x}{\cos x} \right)$$

Berdasarkan aturan pembagian pada turunan, diperoleh $$\begin{aligned} D_x \tan x &= D_x \left( \frac{\textcolor{blue}{\sin x}}{\textcolor{green}{\cos x}} \right) \\ &= \frac{D_x \textcolor{blue}{\sin x} \cdot \textcolor{green}{\cos x} - \textcolor{blue}{\sin x} \cdot D_x \textcolor{green}{\cos x}}{(\textcolor{green}{\cos x})^2} \\ &= \frac{\cos x \cdot \cos x - \sin x (-\sin x)}{\cos^2 x} \\ &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\ &= \frac{1}{\cos^2 x} \\ &= \sec^2 x \end{aligned}$$

Pembahasan
✶ Nomor 4

Tentukan hasil dari $D_x \csc x$.

Pertama, nyatakan $\csc x$ sebagai kebalikan dari $\sin x$. $$D_x \csc x = D_x \left( \frac{1}{\sin x} \right)$$

Berdasarkan aturan pembagian pada turunan, diperoleh $$\begin{aligned} D_x \csc x &= D_x \left( \frac{\textcolor{blue}{1}}{\textcolor{green}{\sin x}} \right) \\ &= \frac{D_x \textcolor{blue}{1} \cdot \textcolor{green}{\sin x} - \textcolor{blue}{1} \cdot D_x \textcolor{green}{\sin x}}{(\textcolor{green}{\sin x})^2} \\ &= \frac{0 \cdot \sin x - 1 \cdot \cos x}{\sin^2 x} \\ &= \frac{0-\cos x}{\sin^2 x} \\ &= \frac{-\cos x}{\sin x \cdot \sin x} \\ &= - \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} \\ &= - \csc x \cdot \cot x \end{aligned}$$

Pembahasan
✶ Nomor 5

Tentukan hasil dari $D_x \sec x$.

Pertama, nyatakan $\sec x$ sebagai kebalikan dari $\cos x$. $$D_x \sec x = D_x \left( \frac{1}{\cos x} \right)$$

Berdasarkan aturan pembagian pada turunan, diperoleh $$\begin{aligned} D_x \sec x &= D_x \left( \frac{\textcolor{blue}{1}}{\textcolor{green}{\cos x}} \right) \\ &= \frac{D_x \textcolor{blue}{1} \cdot \textcolor{green}{\cos x} - \textcolor{blue}{1} \cdot D_x \textcolor{green}{\cos x}}{(\textcolor{green}{\cos x})^2} \\ &= \frac{0 \cdot \cos x - 1 \cdot (- \sin x)}{\cos^2 x} \\ &= \frac{0+\sin x}{\cos^2 x} \\ &= \frac{\sin x}{\cos x \cdot \cos x} \\ &= \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \\ &= \sec x \cdot \tan x \end{aligned}$$

Pembahasan
✶ Nomor 6

Tentukan hasil dari $D_x \cot x$.

Pertama, nyatakan $\cot x$ sebagai hasil bagi antara $\cos x$ dan $\sin x$. $$D_x \cot x = D_x \left( \frac{\cos x}{\sin x} \right)$$

Berdasarkan aturan pembagian pada turunan, diperoleh $$\begin{aligned} D_x \cot x &= D_x \left( \frac{\textcolor{blue}{\cos x}}{\textcolor{green}{\sin x}} \right) \\ &= \frac{D_x \textcolor{blue}{\cos x} \cdot \textcolor{green}{\sin x} - \textcolor{blue}{\cos x} \cdot D_x \textcolor{green}{\sin x}}{(\textcolor{green}{\sin x})^2} \\ &= \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} \\ &= \frac{-\sin^2 x-\cos^2 x}{\sin^2 x} \\ &= \frac{-(\sin^2 x+\cos^2 x)}{\sin^2 x} \\ &= \frac{-1}{\sin^2 x} \\ &= -\csc^2 x \end{aligned}$$

Pembahasan
✶ Nomor 7

Tentukan $D_xy$, jika diketahui $y=2\sin x +3 \cos x$.

Berdasarkan aturan penjumlahan pada turunan, diperoleh $$\begin{aligned} D_xy &= D_x(\textcolor{red}{2\sin x}+\textcolor{blue}{3\cos x}) \\ &= D_x(\textcolor{red}{2 \sin x})+D_x(\textcolor{blue}{3\cos x}) \\ &= 2\cdot D_x \sin x+3 \cdot D_x \cos x \\ &= 2 \cdot \cos x + 3 \cdot (-\sin x) \\ &= 2\cos x-3\sin x \end{aligned}$$

Pembahasan
✶ Nomor 8

Tentukan $D_xy$, jika diketahui $y= \sin^2 x$.

Misalkan $u = \sin x$, sehingga $y=u^2$. Turunan dari kedua fungsi ini adalah $$\begin{aligned} &u = \sin x &&\Longrightarrow \quad \frac{du}{dx} = \cos x \\ &y = u^2 &&\Longrightarrow \quad \frac{dy}{du} = 2u \end{aligned}$$

Berdasarkan Aturan Rantai diperoleh $$\begin{aligned} D_xy &= \frac{dy}{dx} \\ &= \frac{dy}{du} \cdot \frac{du}{dx} \\ &= 2 \textcolor{blue}{u} \cdot \cos x \\ &= 2 \textcolor{blue}{\sin x} \cos x \end{aligned}$$

Pembahasan
✶ Nomor 9

Tentukan $D_xy$, jika diketahui $y= \cos^2 x + \sin^2 x$.

Berdasarkan aturan penjumlahan, diperoleh $$\begin{aligned} D_xy &= D_x(\cos^2 x + \sin^2 x) \\ &= \textcolor{red}{D_x(\cos^2 x)} + \textcolor{blue}{D_x (\sin^2 x)} \end{aligned}$$

Hasil dari $\textcolor{red}{D_x(\cos^2 x)}$ dan $\textcolor{blue}{D_x (\sin^2 x)}$ dapat dihitung menggunakan Aturan Rantai. $$\begin{aligned} D_xy &= \textcolor{red}{2 \cos x (-\sin x)} + \textcolor{blue}{2\sin x \cos x} \\ &= -2\sin x\cos x + 2 \sin x \cos x \\ &= 0 \end{aligned}$$

Cara yang lebih mudah adalah memanfaatkan identitas trigonometri $\cos^2x+\sin^2x=1$. $$\begin{aligned} D_xy &= D_x (\textcolor{teal}{\cos^2 x + \sin^2 x}) \\ &= D_x (\textcolor{teal}{1}) \\ &= 0 \end{aligned}$$

Pembahasan
✶ Nomor 10

Tentukan $D_xy$, jika diketahui $y= 1-\sin^2 x$.

Berdasarkan aturan pengurangan, diperoleh $$\begin{aligned} D_xy &= D_x(1-\sin^2 x) \\ &= \textcolor{red}{D_x(1)}-\textcolor{blue}{D_x (\sin^2 x)} \\ &= \textcolor{red}{0}-\textcolor{blue}{2\sin x\cos x} \\ &= -2\sin x\cos x \end{aligned}$$

Pembahasan
✶ Nomor 11

Tentukan $D_xy$, jika diketahui $y= \frac{\sin x+\cos x}{\cos x}$.

Berdasarkan aturan pembagian, diperoleh $$\begin{aligned} D_xy &= D_x \left( \frac{\textcolor{blue}{\sin x+\cos x}}{\textcolor{green}{\cos x}} \right) \\ &= \frac{D_x (\textcolor{blue}{\sin x+\cos x}) \cdot \textcolor{green}{\cos x} - (\textcolor{blue}{\sin x+\cos x}) \cdot D_x \textcolor{green}{\cos x}}{(\textcolor{green}{\cos x})^2} \\ &= \frac{(\cos x-\sin x) \cdot \cos x-(\sin x+\cos x)(-\sin x)}{\cos^2 x} \\ &= \frac{\cos^2 x-\textcolor{red}{\sin x\cos x} + \sin^2 x + \textcolor{red}{\sin x\cos x}}{\cos^2x} \\ &= \frac{\cos^2x+\sin^2x}{\cos^2x} \\ &= \frac{1}{\cos^2 x} \\ &= \sec^2x \end{aligned}$$

Pembahasan
✶ Nomor 12

Tentukan $D_xy$, jika diketahui $y= \sin x\cos x$.

Berdasarkan aturan perkalian, diperoleh $$\begin{aligned} D_xy &= D_x (\textcolor{red}{\sin x}\textcolor{blue}{\cos x}) \\ &= D_x (\textcolor{red}{\sin x}) \cdot \textcolor{blue}{\cos x} + \textcolor{red}{\sin x} \cdot D_x (\textcolor{blue}{\cos x}) \\ &= \cos x \cdot \cos x + \sin x \cdot (-\sin x) \\ &= \cos^2 x-\sin^2 x \end{aligned}$$

Pembahasan
✶ Nomor 13

Tentukan $D_xy$, jika diketahui $y= \sin x\tan x$.

Berdasarkan aturan perkalian, diperoleh $$\begin{aligned} D_xy &= D_x (\textcolor{red}{\sin x} \textcolor{blue}{\tan x}) \\ &= D_x (\textcolor{red}{\sin x}) \cdot \textcolor{blue}{\tan x} + \textcolor{red}{\sin x} \cdot D_x (\textcolor{blue}{\tan x}) \\ &= \cos x \cdot \tan x + \sin x \cdot \sec^2 x \\ &= \cos x \cdot \frac{\sin x}{\cos x} + \sin x \cdot \frac{1}{\cos^2 x} \\ &= \sin x+\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \\ &= \sin x + \tan x \sec x \end{aligned}$$

Pembahasan
✶ Nomor 14

Tentukan $D_xy$, jika diketahui $y= \frac{\sin x}{x}$.

Berdasarkan aturan pembagian, diperoleh $$\begin{aligned} D_xy &= D_x \left( \frac{\textcolor{red}{\sin x}}{\textcolor{blue}{x}} \right) \\ &= \frac{D_x (\textcolor{red}{\sin x}) \cdot \textcolor{blue}{x}-\textcolor{red}{\sin x} \cdot D_x \textcolor{blue}{x}}{\textcolor{blue}{x}^2} \\ &= \frac{\cos x \cdot x-\sin x \cdot 1}{x^2} \\ &= \frac{x\cos x-\sin x}{x^2} \end{aligned}$$

Pembahasan
✶ Nomor 15

Tentukan $D_xy$, jika diketahui $y= x^2 \cos x$.

Berdasarkan aturan perkalian, diperoleh $$\begin{aligned} D_xy &= D_x (\textcolor{red}{x^2} \textcolor{blue}{\cos x}) \\ &= D_x (\textcolor{red}{x^2}) \cdot \textcolor{blue}{\cos x} + \textcolor{red}{x^2} \cdot D_x (\textcolor{blue}{\cos x}) \\ &= 2x \cdot \cos x + x^2 \cdot (-\sin x) \\ &= 2x\cos x-x^2\sin x \end{aligned}$$

Pembahasan
✶ Nomor 16

Tentukan $D_xy$, jika diketahui $y= \tan^2 x$.

Berdasarkan aturan rantai, diperoleh $$\begin{aligned} D_xy &= D_x (\tan^2 x) \\ &= 2\tan x \cdot \textcolor{blue}{D_x (\tan x)} \\ &= 2\tan x \cdot \textcolor{blue}{\sec^2 x} \end{aligned}$$

Pembahasan
✶ Nomor 17

Tentukan $D_xy$, jika diketahui $y= \sec^3 x$.

Berdasarkan aturan rantai, diperoleh $$\begin{aligned} D_xy &= D_x (\sec^3 x) \\ &= 3\sec^2 x \cdot \textcolor{blue}{D_x (\sec x)} \\ &= 3\sec^2 x \cdot \textcolor{blue}{\sec x \tan x} \\ &= 3\sec^3 x \tan x \end{aligned}$$

Pembahasan
✶ Nomor 18

Gunakan identitas trigonometri $\sin 2x = 2\sin x\cos x$ dan aturan perkalian, untuk menentukan $D_x \sin 2x$.

Berdasarkan identitas trigonometri $\sin 2x = 2\sin x\cos x$ dan aturan perkalian, diperoleh $$\begin{aligned} D_x \sin 2x &= D_x (2\sin x\cos x) \\ &= 2 \cdot D_x (\textcolor{red}{\sin x}\textcolor{blue}{\cos x}) \\ &= 2 \cdot [D_x\textcolor{red}{\sin x} \cdot \textcolor{blue}{\cos x} + \textcolor{red}{\sin x} \cdot D_x \textcolor{blue}{\cos x}] \\ &= 2 \cdot [\cos x \cdot \cos x + \sin x \cdot (-\sin x)] \\ &= 2 \cdot [\cos^2 x-\sin^2 x] \\ &= 2 \cos 2x \end{aligned}$$

Pembahasan