Question

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

Hint:

Use logarithmic differentiation separately on each term, since both base and exponent involve x.

Answer 1

Step 1: Splitting into two parts:
Let \(u=x^{\sin x}\) and \(v=(\cos x)^{\tan x}\), so \(y=u+v\) and \(\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}\).

Step 2: Differentiating u using logarithmic differentiation:
\(\ln u=\sin x\ln x\). Differentiating: \(\dfrac{1}{u}\dfrac{du}{dx}=\cos x\ln x+\dfrac{\sin x}{x}\), so \(\dfrac{du}{dx}=x^{\sin x}\Big(\cos x\ln x+\dfrac{\sin x}{x}\Big)\).

Step 3: Differentiating v using logarithmic differentiation:
\(\ln v=\tan x\ln(\cos x)\). Differentiating: \(\dfrac{1}{v}\dfrac{dv}{dx}=\sec^{2}x\ln(\cos x)+\tan x\cdot\dfrac{-\sin x}{\cos x}=\sec^{2}x\ln(\cos x)-\tan^{2}x\), so \(\dfrac{dv}{dx}=(\cos x)^{\tan x}\big[\sec^{2}x\ln(\cos x)-\tan^{2}x\big]\).

Final Answer:
\[ \boxed{\dfrac{dy}{dx}=x^{\sin x}\Big(\cos x\ln x+\dfrac{\sin x}{x}\Big)+(\cos x)^{\tan x}\big[\sec^{2}x\ln(\cos x)-\tan^{2}x\big]} \]