SBMPTN Zone: Turunan Fungsi Trigonometri

Berikut ini adalah kumpulan soal mengenai turunan fungsi trigonometri tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN

No. 1

Misalkan f(x)=\sin(\sin x\cdot\cos x), maka f'(x)= ....

SBMPTN 2017

$\begin{array}{rl}f(x)& =\sin (\sin x\cdot \cos x)\\ & =\sin ({\displaystyle \frac{1}{2}}\sin 2x)\\ f^{\prime }(x)& =\cos ({\displaystyle \frac{1}{2}}\sin 2x)\cdot \cos 2x\\ & =\cos 2x\cdot \cos ({\displaystyle \frac{1}{2}}\sin 2x)\end{array}$

Jadi, f'(x)=\cos2x\cdot\cos\left(\dfrac12\sin2x\right).

No. 2

Jika f(x)=\sqrt{\sin x+\sqrt{\cos x+\sqrt{\sin x}}}, maka f'(x)=...

$\begin{array}{rl}{f}^{\prime }(x)& ={\displaystyle \frac{\cos x+{\displaystyle \frac{-\sin x+{\displaystyle \frac{\cos x}{2\sqrt{\sin x}}}}{2\sqrt{\cos x+\sqrt{\sin x}}}}\cdot {\displaystyle \frac{2\sqrt{\sin x}}{2\sqrt{\sin x}}}}{2\sqrt{\sin x+\sqrt{\cos x+\sqrt{\sin x}}}}}\\ & ={\displaystyle \frac{\cos x+{\displaystyle \frac{-2\sin x\sqrt{\sin x}+\cos x}{4\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}}}{2\sqrt{\sin x+\sqrt{\cos x+\sqrt{\sin x}}}}}\cdot {\displaystyle \frac{4\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}{4\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}}\\ & ={\displaystyle \frac{4\cos x\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}-2\sin x\sqrt{\sin x}+\cos x}{8\sqrt{{\sin }^{2}x\cos x+{\sin }^{2}x\sqrt{\sin x}+\sin x\cos x\sqrt{\cos x+\sqrt{\sin x}}+\sin x\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}}}\end{array}$

Jadi, f'(x)=\dfrac{4\cos x\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}-2\sin x\sqrt{\sin x}+\cos x}{8\sqrt{\sin^2x\cos x+\sin^2x\sqrt{\sin x}+\sin x\cos x\sqrt{\cos x+\sqrt{\sin x}}+\sin x\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}}.

No. 3

Jika {f(x) = \sin (\cos 2x)}, maka {f'(x) =}

1. -2\sin 2x \sin(\sin 2x)
2. 2\sin 2x\sin(\cos 2x)
3. -2sin 2x \sin(\cos 2x)

4. 2\sin 2x \cos(\cos 2x)
5. -2\sin 2x \cos(\cos 2x)

SKMP September 2020

$\begin{array}{rl}f(x)& =\sin (\cos 2x)\\ f^{\prime }(x)& =\boxed{{\displaystyle \boxed{{\displaystyle -2\sin 2x\cos (\cos 2x)}}}}\end{array}$

Jadi, {f'(x) =-2\sin2x\cos(\cos2x)}
JAWAB E