Find the derivative of $tanx$ $cosx$

The correct answer is $cosx$

Let f(x) = $tanx$ $cosx$

Using the Product rule,

$f'(x)$ = $tanx \dfrac{d}{dx}cosx$ + $cosx \dfrac{d}{dx}tanx$

= $tanx (-sinx)$ + $cosx (sec^2x)$

= $-sinx$ $tanx$ + $(sec^2x) cosx$

= $\dfrac{-sin^2}{cosx}$ + $\dfrac{1}{cosx}$

= $\dfrac{1-sin^2x}{cosx}$

= $\dfrac{cos^2x}{cosx}$

= $cosx$