Answer
$\sec^2 x-\cos x$
Work Step by Step
We have: $f(x)=\tan x-\sin x$
We need to differentiate both sides with respect to $x$.
$\dfrac{d[f(x)]}{dx}= \dfrac{d(\tan x-\sin x)}{dx} \\ f^{\prime}(x)= \dfrac{d}{dx}(\tan x)- \dfrac{d}{dx}(\sin x)$
Therefore, $f^{\prime}(x)=\sec^2 x-\cos x$
