Question

If $$ y = (\log_x \sin x)^x, $$ then find $$ \frac{dy}{dx}. $$

• A. \( y \left[ \frac{x \sin x}{\log x \cos x} + \frac{1}{\log x} - \log(\log x) \right] \)
• B. \( y \left[ \frac{x \cos x}{\log \sin x} - \log(\log \sin x) + \frac{1}{\log x} \right] \)
• C. \( y \left[ \frac{x \cot x}{\log \sin x} + \log(\log \sin x) - \frac{1}{\log x} \right] \)
• D. \( y \left[ \frac{x \cot x}{\log \sin x} - \log(\log \sin x) + \frac{1}{\log x} \right] \)

Hint:

Use logarithmic differentiation for functions with variable bases and exponents.

Answer 1

The Correct Option is C

Write \[ y = \left( \frac{\log \sin x}{\log x} \right)^x = e^{x \log \left( \frac{\log \sin x}{\log x} \right)} \] Differentiate: \[ \frac{dy}{dx} = y \left[ \log \left( \frac{\log \sin x}{\log x} \right) + x \frac{d}{dx} \log \left( \frac{\log \sin x}{\log x} \right) \right] \] Calculate derivative inside bracket using chain rule: \[ \frac{d}{dx} \log \left( \frac{\log \sin x}{\log x} \right) = \frac{1}{\log \sin x} \cdot \frac{\cos x}{\sin x} - \frac{1}{\log x} \cdot \frac{1}{x} \] After simplification: \[ \frac{dy}{dx} = y \left[ \frac{x \cot x}{\log \sin x} + \log(\log \sin x) - \frac{1}{\log x} \right] \]