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

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Answer 1 · 2016-07-19T01:47:46

$f'[g(x)] = -2e^x(e^x+1)$

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

Replacing $x$ by $g(x)$ in $f(x)$

$f[g(x)] = -(e^x)^2 -2*e^x$

$f' [g(x)] = d/dx(-(e^x)^2 -2*e^x) = = d/dx(-e^(2x) -2*e^x)$

$= -2e^(2x)-2e^x$

$= -2e^x(e^x+1)$

If f(x) = -x^2 -2xf(x)=−x2−2xf(x) = -x^2 -2x and g(x) = e^(x)g(x)=exg(x) = e^(x), what is f'(g(x)) f'(g(x)) ?