If $f (x) = - e^{x}$ $f(x) =-e^(x)$ and $g (x) = {tan}^{2} x^{2}$ $g(x) = tan^2x^2$ , what is $f'(g(x))$ ?

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Answer 1 · 2017-04-16T00:12:55

First, take the derivative of $f(x)$ .

$f'(x)=-e^x$

Note that the derivative of $e^x$ is just $e^x$ and adding a negative to it won't change that.

Now, put in $g(x)$ for where there's a $x$ in $f(x)$

$f'(g(x))=-e^(tan^2(x^2))$

If f(x) =-e^(x) f(x)=−exf(x) =-e^(x) and g(x) = tan^2x^2 g(x)=tan2x2g(x) = tan^2x^2 , what is f'(g(x)) f'(g(x)) ?