# How do you differentiate e^((x^2-1)^2)  using the chain rule?

Oct 21, 2017

$\frac{d}{\mathrm{dx}} {e}^{{\left({x}^{2} - 1\right)}^{2}} = 4 {e}^{{\left({x}^{2} - 1\right)}^{2}} \cdot x \left({x}^{2} - 1\right)$

#### Explanation:

$y = {e}^{{\left({x}^{2} - 1\right)}^{2}}$

Let $u = {\left({x}^{2} - 1\right)}^{2}$ such that $y = {e}^{u}$

Now $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

To get the derivative of $u$, we use the chain rule again:

Let $v = {x}^{2} - 1$ such that $u = {v}^{2}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = 2 x$

$\frac{\mathrm{du}}{\mathrm{dv}} = 2 v$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{dx}} \cdot \frac{\mathrm{du}}{\mathrm{dv}} = 4 x v = 4 x \left({x}^{2} - 1\right)$

$\frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u} = {e}^{{\left({x}^{2} - 1\right)}^{2}}$

Now we can go back to the original equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = {e}^{{\left({x}^{2} - 1\right)}^{2}} \cdot 4 x \left({x}^{2} - 1\right)$

In the answers, variables $v$ and $u$ were defined for the use of chain rule.