# If f(x)= - e^x  and g(x) = sqrt(1-x^2 , how do you differentiate f(g(x))  using the chain rule?

Jul 18, 2017

$f ' \left(g \left(x\right)\right) = \frac{x {e}^{\sqrt{1 - {x}^{2}}}}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

$f \left(x\right) = - {e}^{x}$ and $g \left(x\right) = \sqrt{1 - {x}^{2}}$

:. f(g(x)) = -e^(sqrt(1-x^2)

Applying the chain rule:

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

Applying the chain rule agsin:

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

Applying the chain rule agsin:

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

$= \frac{x {e}^{\sqrt{1 - {x}^{2}}}}{\sqrt{1 - {x}^{2}}}$