If f(x) =xe^x and g(x) = sinx-x, what is f'(g(x)) ?

Jul 11, 2018

$f ' \left(g \left(x\right)\right) = \left(\cos x - 1\right) {e}^{\sin x - x} \left(\sin x - x + 1\right)$

Explanation:

$f \left(x\right) = x {e}^{x}$
$g \left(x\right) = \sin x - x$

$f \left(g \left(x\right)\right)$ means that we sub $\sin x - x$ into any $x$ in $f \left(x\right)$
$f \left(g \left(x\right)\right) = \left(\sin x - x\right) {e}^{\sin x - x}$

$f ' \left(g \left(x\right)\right) = \left(\sin x - x\right) \times \left(\cos x - 1\right) {e}^{\sin x - x} + {e}^{\sin x - x} \times \left(\cos x - 1\right)$

$f ' \left(g \left(x\right)\right) = \left(\sin x - x\right) \left(\cos x - 1\right) {e}^{\sin x - x} + \left(\cos x - 1\right) {e}^{\sin x - x}$

$f ' \left(g \left(x\right)\right) = \left(\cos x - 1\right) {e}^{\sin x - x} \left(\sin x - x + 1\right)$