MCQ
The derivative of $\sin ^2 x$ w.r.t. $e^{\cos x} i$
- A$\frac{2}{e^{\cos x}}$
- B$\frac{2 \cos x}{e^{\cos x}}$
- ✓$-\frac{2 \cos x}{e^{\cos x}}$
- D$\frac{e^{\cos x}}{-2}$
✓
Answer
Correct option: C.
$-\frac{2 \cos x}{e^{\cos x}}$
(c) $-\frac{2 \cos x}{e^{\cos x}}$
Explanation: Let $u(x)=\sin ^2 x$ and $v(x)=e^{\cos x}$.
We want to find $\frac{d u}{d v}=\frac{\frac{d u}{d z}}{\frac{d v}{d x}}$.
Clearly, $\frac{d u}{d x}=2 \sin x \cos x$ and $\frac{d v}{d x}= e ^{\cos x }(-\sin x )=-(\sin x ) e ^{\cos x }$ $\frac{d u}{d v}=\frac{2 \sin x \cos x}{-\sin x e^{\cos \alpha}}=-\frac{2 \cos x}{e^{\cos \alpha}}$
Explanation: Let $u(x)=\sin ^2 x$ and $v(x)=e^{\cos x}$.
We want to find $\frac{d u}{d v}=\frac{\frac{d u}{d z}}{\frac{d v}{d x}}$.
Clearly, $\frac{d u}{d x}=2 \sin x \cos x$ and $\frac{d v}{d x}= e ^{\cos x }(-\sin x )=-(\sin x ) e ^{\cos x }$ $\frac{d u}{d v}=\frac{2 \sin x \cos x}{-\sin x e^{\cos \alpha}}=-\frac{2 \cos x}{e^{\cos \alpha}}$
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