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.