Question

Find the differential coefficient of \[ y = x^x + (\cos x)^{\tan x}. \]

Hint:

For differentiating expressions like \( x^x \), use logarithmic differentiation, and for \( (\cos x)^{\tan x} \), apply the chain rule after logarithmic differentiation.

Answer 1

We are tasked with finding \( \frac{dy}{dx} \). The given function is: \[ y = x^x + (\cos x)^{\tan x} \] 1. Step 1: Differentiating \( x^x \) To differentiate \( x^x \), we take the natural logarithm of both sides: \[ \ln y = \ln (x^x) = x \ln x \] Now, differentiate implicitly with respect to \( x \): \[ \frac{d}{dx} \ln y = \frac{d}{dx} (x \ln x) \] \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \] Thus: \[ \frac{dy}{dx} = y (\ln x + 1) \] Substitute \( y = x^x \) back: \[ \frac{dy}{dx} = x^x (\ln x + 1) \] 2. Step 2: Differentiating \( (\cos x)^{\tan x} \) Let \( z = (\cos x)^{\tan x} \). We differentiate using logarithmic differentiation: \[ \ln z = \tan x \ln (\cos x) \] Now, differentiate implicitly: \[ \frac{d}{dx} \ln z = \frac{d}{dx} (\tan x \ln (\cos x)) \] Apply the product rule: \[ \frac{1}{z} \frac{dz}{dx} = \sec^2 x \ln (\cos x) + \tan x \cdot \frac{-\sin x}{\cos x} \] Thus: \[ \frac{dz}{dx} = (\cos x)^{\tan x} \left( \sec^2 x \ln (\cos x) - \tan x \cdot \tan x \right) \] 3. Step 3: Final Expression Now, the total derivative of \( y \) is: \[ \frac{dy}{dx} = x^x (\ln x + 1) + (\cos x)^{\tan x} \left( \sec^2 x \ln (\cos x) - \tan^2 x \right) \] Final Answer: \[ \boxed{\frac{dy}{dx} = x^x (\ln x + 1) + (\cos x)^{\tan x} \left( \sec^2 x \ln (\cos x) - \tan^2 x \right)} \]