Question

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

Answer 1

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

Explanation:

`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.