If $f (x) = - e^{- 3 x - 7}$ $f(x) =-e^(-3x-7)$ and $g (x) = {ln x}^{2}$ $g(x) = lnx^2$ , what is $f'(g(x))$ ?

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Answer 1 · 2017-07-28T14:38:50

If $f(x) =-e^(-3x-7)$ and $g(x) = lnx^2$ , what is $f'(g(x))$ ?

Given

$f(x) =-e^(-3x-7)$

Differentiating w . r to x we gat

$f'(x) =-e^(-3x-7)xx(d/(dx) (-3x-7))$

$=>f'(x) =3e^(-3x-7)$

Inserting $x=g(x)$ we have

$f'(g(x)) =3e^(-3g(x)-7)$

$=>f'(g(x)) =3e^-7xxe^(-3g(x))$

$=>f'(g(x)) =3e^-7xxe^(-3 lnx^2)$

$=>f'(g(x)) =3e^-7xxe^( lnx^(2*(-3))$

$=>f'(g(x)) =3e^-7xxe^( lnx^(-6))$

$=>f'(g(x)) =3e^-7xxx^(-6)$

If f(x) =-e^(-3x-7) f(x)=−e−3x−7f(x) =-e^(-3x-7) and g(x) = lnx^2 g(x)=lnx2g(x) = lnx^2 , what is f'(g(x)) f'(g(x)) ?