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

\(\eqalign{ f(x)&=\sin(\sin x\cdot\cos x)\\ &=\sin\left(\dfrac12\sin2x\right)\\ f'(x)&=\cos\left(\dfrac12\sin2x\right)\cdot\cos2x\\ &=\cos2x\cdot\cos\left(\dfrac12\sin2x\right) }\)

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)=...

\(\begin{aligned} f'(x)&=\dfrac{\cos x+\dfrac{-\sin x+\dfrac{\cos x}{2\sqrt{\sin x}}}{2\sqrt{\cos x+\sqrt{\sin x}}}\cdot\color{red}{\dfrac{2\sqrt{\sin x}}{2\sqrt{\sin x}}}}{2\sqrt{\sin x+\sqrt{\cos x+\sqrt{\sin x}}}}\\[8pt] &=\dfrac{\cos x+\dfrac{-2\sin x\sqrt{\sin x}+\cos x}{4\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}}{2\sqrt{\sin x+\sqrt{\cos x+\sqrt{\sin x}}}}\cdot\color{red}{\dfrac{4\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}{4\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}}\\[8pt] &=\dfrac{4\cos x\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}-2\sin x\sqrt{\sin x}+\cos x}{8\sqrt{\sin^2x\cos x+\sin^2x\sqrt{\sin x}+\sin x\cos x\sqrt{\cos x+\sqrt{\sin x}}+\sin x\sqrt{\sin x\cos x+\sin x\sqrt{\sin x}}}} \end{aligned}\)

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

\(\eqalign{ f(x) &= \sin (\cos 2x)\\ f'(x)&=\boxed{\boxed{-2\sin2x\cos(\cos2x)}} }\)

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