Question Detail

This is a logarithmic differentiation problem since both the base and the exponent are functions of $x$.Given $y = x^{\tan x}$.Take the natural logarithm of both sides:$\ln y = \ln (x^{\tan x})$Use the logarithm property $\ln(a^b) = b\ln(a)$:$\ln y = \tan x \cdot \ln x$Now, differentiate both sides implicitly with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\tan x \cdot \ln x)$Apply the product rule on the right side:$\frac{d}{dx}(\tan x) \cdot \ln x + \tan x \cdot \frac{d}{dx}(\ln x)$We know $\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(\ln x) = \frac{1}{x}$:$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \ln x + \frac{\tan x}{x}$Finally, multiply both sides by $y$ and substitute $y = x^{\tan x}$ back:$\frac{dy}{dx} = y \left(\sec^2 x \ln x + \frac{\tan x}{x}\right)$$\frac{dy}{dx} = x^{\tan x} \left(\frac{\tan x}{x} + \sec^2 x \ln x\right)$This matches option D.