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

$f ' \left(g \left(x\right)\right) = 4 x \cdot {e}^{- {\tan}^{2} {x}^{2}} \cdot \left({\tan}^{3} {x}^{2} + \tan {x}^{2}\right)$

Explanation:

given $f \left(x\right) = - {e}^{- x}$ and $g \left(x\right) = {\tan}^{2} {x}^{2}$

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

$f \left(g \left(x\right)\right) = - {e}^{- {\tan}^{2} {x}^{2}}$

$f ' \left(g \left(x\right)\right) = \left(- 1\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{- {\tan}^{2} {x}^{2}}\right)$

$f ' \left(g \left(x\right)\right) = \left(- 1\right) \cdot \left({e}^{- {\tan}^{2} {x}^{2}}\right) \cdot \frac{d}{\mathrm{dx}} \left(- {\tan}^{2} {x}^{2}\right)$

$f ' \left(g \left(x\right)\right) = \left(- 1\right) \cdot \left({e}^{- {\tan}^{2} {x}^{2}}\right) \cdot - \frac{d}{\mathrm{dx}} \left({\tan}^{2} {x}^{2}\right)$

$f ' \left(g \left(x\right)\right) = \left(- 1\right) \cdot \left({e}^{- {\tan}^{2} {x}^{2}}\right) \left(- 2 {\left(\tan {x}^{2}\right)}^{2 - 1}\right) \frac{d}{\mathrm{dx}} \left(\tan {x}^{2}\right)$

$f ' \left(g \left(x\right)\right) = \left(- 1\right) \cdot \left({e}^{- {\tan}^{2} {x}^{2}}\right) \left(- 2 {\left(\tan {x}^{2}\right)}^{2 - 1}\right) \left({\sec}^{2} {x}^{2}\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

$f ' \left(g \left(x\right)\right) = \left(- 1\right) \cdot \left({e}^{- {\tan}^{2} {x}^{2}}\right) \left(- 2 {\left(\tan {x}^{2}\right)}^{2 - 1}\right) \left({\sec}^{2} {x}^{2}\right) \left(2 x\right)$

Simplify at this point

$f ' \left(g \left(x\right)\right) = \left(4 x\right) \cdot \left({e}^{- {\tan}^{2} {x}^{2}}\right) \left(\tan {x}^{2}\right) \left({\sec}^{2} {x}^{2}\right)$

$f ' \left(g \left(x\right)\right) = \left(4 x\right) \cdot \left({e}^{- {\tan}^{2} {x}^{2}}\right) \left(\tan {x}^{2}\right) \left({\tan}^{2} {x}^{2} + 1\right)$

$f ' \left(g \left(x\right)\right) = \left(4 x\right) \cdot \left({e}^{- {\tan}^{2} {x}^{2}}\right) \left({\tan}^{3} {x}^{2} + \tan {x}^{2}\right)$

God bless....I hope the explanation is useful.