Question

Differentiate $y = (\cos x)^{\tan x} + x^{x}$.

Hint:

For $f(x)^{g(x)}$, use $\ln$ differentiation: $\ln y = g(x)\ln f(x)$.

Answer 1

We have two terms: $y = (\cos x)^{\tan x} + x^{x}$ --- Part 1: Differentiate $(\cos x)^{\tan x$} Let $u = (\cos x)^{\tan x}$ \[ \ln u = \tan x \cdot \ln(\cos x) \] Differentiate w.r.t $x$: \[ \frac{1}{u}\frac{du}{dx} = \sec^{2}x \cdot \ln(\cos x) + \tan x \cdot \frac{1}{\cos x}(-\sin x) \] \[ \frac{1}{u}\frac{du}{dx} = \sec^{2}x \cdot \ln(\cos x) - \tan^{2}x \] So, \[ \frac{du}{dx} = (\cos x)^{\tan x}\left(\sec^{2}x \cdot \ln(\cos x) - \tan^{2}x\right) \] --- Part 2: Differentiate $x^{x$} Let $v = x^{x}$ \[ \ln v = x\ln x \] Differentiate: \[ \frac{1}{v}\frac{dv}{dx} = \ln x + 1 \] \[ \frac{dv}{dx} = x^{x}(\ln x + 1) \] --- Final derivative: \[ \frac{dy}{dx} = (\cos x)^{\tan x}\left(\sec^{2}x \cdot \ln(\cos x) - \tan^{2}x\right) + x^{x}(\ln x + 1) \] Final Answer: \[ \boxed{\dfrac{dy}{dx} = (\cos x)^{\tan x}\Big(\sec^{2}x \cdot \ln(\cos x) - \tan^{2}x\Big) + x^{x}(\ln x + 1)} \]