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

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Answer 1 · 2017-10-21T12:39:32

$d/dx e^((x^2-1)^2) = 4e^((x^2-1)^2)*x(x^2-1)$

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

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

Now $dy/dx = dy/(du) * (du)/(dx)$

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

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

$(dv)/(dx) = 2x$

$(du)/(dv) = 2v$

$(du)/(dx) = (dv)/(dx) * (du)/(dv) = 4xv = 4x(x^2-1)$

$dy/(du) = e^u = e^((x^2-1)^2)$

Now we can go back to the original equation:

$dy/dx = dy/(du) * (du)/(dx) = e^((x^2-1)^2) * 4x(x^2-1)$

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

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