Question

If \( y = (\cot x)^{\sin x} + x^x \), then find \( \frac{dy}{dx} \).

Hint:

For functions like \( (\cot x)^{\sin x} \), use logarithmic differentiation; for \( x^x \), use \( e^{x \ln x} \).

Answer 1

For \( y = (\cot x)^{\sin x} + x^x \): \[ \frac{dy}{dx} = \frac{d}{dx}\left((\cot x)^{\sin x}\right) + \frac{d}{dx}(x^x). \] 1. Differentiate \( (\cot x)^{\sin x} \): Let \( u = \sin x \ln (\cot x) \), then \( (\cot x)^{\sin x} = e^u \). \[ \frac{d}{dx} \left((\cot x)^{\sin x}\right) = e^u \cdot \frac{du}{dx}. \] \[ \frac{du}{dx} = \cos x \ln (\cot x) - \sin x \cdot \frac{\csc^2 x}{\cot x}. \] \[ \frac{d}{dx} \left((\cot x)^{\sin x}\right) = (\cot x)^{\sin x} \left[\cos x \ln (\cot x) - \frac{\sin x}{\cot x}\csc^2 x \right]. \] 2. Differentiate \( x^x \): Using \( x^x = e^{x \ln x} \): \[ \frac{d}{dx}(x^x) = x^x (\ln x + 1). \] Thus: \[ \frac{dy}{dx} = (\cot x)^{\sin x} \left[\cos x \ln (\cot x) - \frac{\sin x}{\cot x}\csc^2 x \right] + x^x (\ln x + 1). \]