If $y = (\sin x)^x$, then $\dfrac{dy}{dx}$ equals:

A$(\sin x)^x [\sin x + x\cos x]$

B$x (\sin x)^{x-1} \cos x$

C$(\sin x)^x \cdot \cos x$

D$(\sin x)^x [\log(\sin x) + x \cot x]$

Answer & Solution

Correct answer: D. $(\sin x)^x [\log(\sin x) + x \cot x]$

**Logarithmic diff:** $\ln y = x \ln(\sin x)$. **Differentiate:** $y'/y = \ln(\sin x) + x \cdot (\cos x/\sin x) = \ln(\sin x) + x\cot x$. **So $y' = y[\ln(\sin x) + x\cot x]$.**

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