Answer
$$f'(x)=\frac{-\sin x}{\ln {10}(1+\cos x)}$$
Work Step by Step
$f'(x)=\frac{d}{dx}\log_{10}(1+\cos x)$
$=\frac{d}{dx}\frac{\ln(1+\cos x)}{\ln 10}$
$=\frac{1}{\ln 10}\frac{d}{dx}\ln (1+\cos x)$
Using the chain rule:
$f'(x)=\frac{1}{\ln 10}[\frac{d\ln(1+\cos x)}{d1+\cos x}
\times \frac{d1+\cos x}{dx}]$
$=\frac{1}{\ln 10}[\frac{1}{1+\cos x}\times-\sin x]$
$=\frac{-\sin x}{\ln {10}(1+\cos x)}$
