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Question
Differentiate \[\left( 1 + \cos x \right)^x\] ?
Solution
\[\text{Let y }= \left( 1 + \cos x \right)^x . . . \left( i \right)\]
Taking log on both sides,
\[\log y = \log \left( 1 + \cos x \right)^x \]
\[ \Rightarrow \log y = x \log\left( 1 + \cos x \right)\]
Differentiating with respect to x,
\[\frac{1}{y}\frac{dy}{dx} = x\frac{d}{dx}\log\left( 1 + \cos x \right) + \log\left( 1 + \cos x \right)\frac{d}{dx}\left( x \right) \]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x \times \frac{1}{\left( 1 + \cos x \right)}\frac{d}{dx}\left( 1 + \cos x \right) + \log\left( 1 + \cos x \right)\left( 1 \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{x}{\left( 1 + \cos x \right)}\left( 0 - \sin x \right) + \log\left( 1 + \cos x \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log\left( 1 + \cos x \right) - \frac{x \sin x}{\left( 1 + \cos x \right)}\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ \log\left( 1 + \cos x \right) - \frac{x \sin x}{\left( 1 + \cos x \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \left( 1 + \cos x \right)^x \left[ \log\left( 1 + \cos x \right) - \frac{x \sin x}{\left( 1 + \cos x \right)} \right] \left[\text{ using equation} \left( i \right) \right]\]
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