alleryStart learning →If $y = (\sin x)^x$, then $\dfrac{dy}{dx}$ equals:
- $(\sin x)^x [\sin x + x\cos x]$
- $x (\sin x)^{x-1} \cos x$
- $(\sin x)^x \cdot \cos x$
- $(\sin x)^x [\log(\sin x) + x \cot x]$
Show answer and 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 $y' = y[\ln(\sin x) + x\cot x]$.**