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

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

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Answer 1 · 2017-10-21T12:39:32

$\frac{d}{d x} e^{{(x^{2} - 1)}^{2}} = 4 e^{{(x^{2} - 1)}^{2}} \cdot x (x^{2} - 1)$ $d/dx e^((x^2-1)^2) = 4e^((x^2-1)^2)*x(x^2-1)$

Explanation:

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

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

Now $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx = dy/(du) * (du)/(dx)$

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

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

$\frac{d v}{d x} = 2 x$ $(dv)/(dx) = 2x$

$\frac{d u}{d v} = 2 v$ $(du)/(dv) = 2v$

$\frac{d u}{d x} = \frac{d v}{d x} \cdot \frac{d u}{d v} = 4 x v = 4 x (x^{2} - 1)$ $(du)/(dx) = (dv)/(dx) * (du)/(dv) = 4xv = 4x(x^2-1)$

$\frac{d y}{d u} = e^{u} = e^{{(x^{2} - 1)}^{2}}$ $dy/(du) = e^u = e^((x^2-1)^2)$

Now we can go back to the original equation:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} = e^{{(x^{2} - 1)}^{2}} \cdot 4 x (x^{2} - 1)$ $dy/dx = dy/(du) * (du)/(dx) = e^((x^2-1)^2) * 4x(x^2-1)$

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

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

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