Question

If \( y = (\sin x)^y \), then \( \frac{dy}{dx} \) is:

• A. \( \frac{y^2 \cot x}{1 - y \log (\sin x)} \)
• B. \( \frac{y^2 \cot x}{1 - y \log (x)} \)
• C. \( \frac{y^2 \cot x}{1 + y \log (\sin x)} \)
• D. \( \frac{y^2 \cot x}{1 + y \log (x)} \)

Hint:

For equations involving \( y \) in both the base and exponent, take the natural logarithm and apply implicit differentiation carefully.

Answer 1

The Correct Option is A

Given: \[ y = (\sin x)^y. \] 
Take the natural logarithm on both sides: \[ \ln y = y \ln (\sin x). \] 
Differentiate both sides with respect to \( x \): \[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \big[ y \ln (\sin x) \big]. \] 
Apply the product rule to the right-hand side: \[ \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \ln (\sin x) + y \frac{d}{dx} \big[ \ln (\sin x) \big]. \] 
The derivative of \( \ln (\sin x) \) is: \[ \frac{d}{dx} \ln (\sin x) = \cot x. \] 
Substitute this into the equation: \[ \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \ln (\sin x) + y \cot x. \] 
Multiply through by \( y \) to eliminate the denominator: \[ \frac{dy}{dx} = y \frac{dy}{dx} \ln (\sin x) + y^2 \cot x. \] 
Rearrange to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \big( 1 - y \ln (\sin x) \big) = y^2 \cot x. \] 
Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{y^2 \cot x}{1 - y \ln (\sin x)}. \] 

Final Answer: \[ \boxed{\frac{y^2 \cot x}{1 - y \ln (\sin x)}} \]