Default
Question
Find the derivative of $tanx$ $cosx$
Solution
The correct answer is $cosx$
Explanation
Let f(x) = $tanx$ $cosx$
Using the Product rule,
$f'(x)$ = $tanx \dfrac{d}{dx}cosx$ + $cosx \dfrac{d}{dx}tanx$
= $tanx (-sinx)$ + $cosx (sec^2x)$
= $-sinx$ $tanx$ + $(sec^2x) cosx$
= $\dfrac{-sin^2}{cosx}$ + $\dfrac{1}{cosx}$
= $\dfrac{1-sin^2x}{cosx}$
= $\dfrac{cos^2x}{cosx}$
= $cosx$
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