Question

Differentiate

${\left(1+\cos x\right)}^x$

Answer 1

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