If \(y = (\sin x)^{y}\), where \(0 \lt x \lt \displaystyle\frac{\pi}{2}\) and \(y \gt 0\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{y^{2}\cot x}{1 - y\log_e(\sin x)}\)
The variable y appears in the exponent, so we use logarithmic differentiation:
\(\log_e y = y\log_e(\sin x)\)
Differentiating both sides with respect to x (implicit differentiation, product rule on the right):
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \frac{dy}{dx}\log_e(\sin x) + y \cdot \frac{\cos x}{\sin x} \)
Collecting the \(\frac{dy}{dx}\) terms:
\( \displaystyle \frac{dy}{dx}\left(\frac{1}{y} - \log_e(\sin x)\right) = y\cot x \)
\( \displaystyle \frac{dy}{dx} \cdot \frac{1 - y\log_e(\sin x)}{y} = y\cot x \)
\( \displaystyle \frac{dy}{dx} = \frac{y^{2}\cot x}{1 - y\log_e(\sin x)} \)
(Here \(\log_e(\sin x) \lt 0\), so the denominator is positive.)