Question Detail

This problem requires logarithmic differentiation because 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 $\ln(a^b) = b\ln a$:$\ln y = (\tan x)(\ln x)$Now, differentiate both sides with respect to $x$ implicitly. On the left, use the chain rule. On the right, use the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\tan x)(\ln x) + (\tan x)\frac{d}{dx}(\ln x)$$\frac{d}{dx}(\tan x) = \sec^2 x$$\frac{d}{dx}(\ln x) = \frac{1}{x}$So,$\frac{1}{y} \frac{dy}{dx} = (\sec^2 x)(\ln x) + (\tan x)\left(\frac{1}{x}\right)$$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \ln x + \frac{\tan x}{x}$Multiply both sides by $y$:$\frac{dy}{dx} = y \left(\sec^2 x \ln x + \frac{\tan x}{x}\right)$Substitute $y = x^{\tan x}$ back into the equation:$\frac{dy}{dx} = x^{\tan x} \left[\frac{\tan x}{x} + \sec^2 x \ln x\right]$