Problem Statement
Find $$\frac{dy}{dx}$$ if $$y = (\sin x)^{\cos x} + (\cos x)^{\sin x}$$.
Verified Solution & Marking Scheme
Split into u and v
Let $$y = u + v$$, so $$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$, where $$u = (\sin x)^{\cos x}$$ and $$v = (\cos x)^{\sin x}$$.
Differentiate u using logarithms
$$\ln u = \cos x \ln(\sin x)$$\n$$\frac{1}{u}\frac{du}{dx} = -\sin x \ln(\sin x) + \cos x \cdot \frac{\cos x}{\sin x} = \cos x \cot x - \sin x \ln(\sin x)$$\n$$\frac{du}{dx} = (\sin x)^{\cos x} [\cos x \cot x - \sin x \ln(\sin x)]$$
Differentiate v using logarithms
$$\ln v = \sin x \ln(\cos x)$$\n$$\frac{1}{v}\frac{dv}{dx} = \cos x \ln(\cos x) + \sin x \cdot \frac{-\sin x}{\cos x} = \cos x \ln(\cos x) - \sin x \tan x$$\n$$\frac{dv}{dx} = (\cos x)^{\sin x} [\cos x \ln(\cos x) - \sin x \tan x]$$
Combine results
$$\frac{dy}{dx} = (\sin x)^{\cos x} [\cos x \cot x - \sin x \ln(\sin x)] + (\cos x)^{\sin x} [\cos x \ln(\cos x) - \sin x \tan x]$$