Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 71 Mark
Question
The derivative of $\sin ^2 x$ w.r.t. $e^{\cos x} i$
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Answer
(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 $\frac{d v}{d x}= e ^{\cos x }(-\sin x )=-(\sin x ) e ^{\cos x }$ $\frac{d u}{d v}=\frac{2 \sin x \cos x}{-\sin x e^{\cos \alpha}}=-\frac{2 \cos x}{e^{\cos \alpha}}$
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