If \(y = x^{\sin x}\), \(x \gt 0\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(x = \displaystyle\frac{\pi}{2}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
Both the base and the exponent contain x, so we use logarithmic differentiation. Taking log on both sides:
\( \log_e y = \sin x\,\log_e x \)
Differentiating, with the product rule on the right:
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \cos x\,\log_e x + \frac{\sin x}{x} \)
\( \displaystyle \frac{dy}{dx} = x^{\sin x}\left(\cos x\,\log_e x + \frac{\sin x}{x}\right) \)
At \(x = \frac{\pi}{2}\), \(\sin x = 1\) and \(\cos x = 0\):
\( \displaystyle \frac{dy}{dx} = \frac{\pi}{2}\left(0 + \frac{1}{\pi/2}\right) \)
\( = 1 \)