Question

If \[ f(x)=\left(\log x\right)^{\sin x},\qquad x>e, \] then \(f'(\pi)=\)

• A. \(\log \pi\)
• B. \(-\log \pi\)
• C. \(\log(\log \pi)\)
• D. \(-\log(\log \pi)\)

Hint:

For functions of the form \[ y=u(x)^{v(x)}, \] use logarithmic differentiation: \[ \boxed{\ln y=v\ln u.} \] Then differentiate implicitly and finally multiply by \(y\).

Answer 1

The Correct Option is D

Step 1: Apply logarithmic differentiation. Given, \[ f(x)=\left(\log x\right)^{\sin x}. \] Taking natural logarithm on both sides, \[ \ln f(x)=\sin x\cdot\ln(\log x). \] Differentiating both sides, \[ \frac{f'(x)}{f(x)} = \cos x\,\ln(\log x) + \sin x\cdot \frac{1}{\log x}\cdot\frac1x. \] Hence, \[ f'(x) = (\log x)^{\sin x} \left[ \cos x\,\ln(\log x) + \frac{\sin x}{x\log x} \right]. \]

Step 2: Substitute \(x=\pi\). Since \[ \sin\pi=0, \qquad \cos\pi=-1, \] we have \[ (\log\pi)^{\sin\pi} = (\log\pi)^0 = 1, \] and \[ \frac{\sin\pi}{\pi\log\pi}=0. \] Therefore, \[ f'(\pi) = 1\left[-\ln(\log\pi)\right]. \] Hence, \[ \boxed{f'(\pi)=-\log(\log\pi).} \] Thus, the correct option is \(\boxed{(D)}\).