How do you differentiate #f(x)=e^(x^2-3x^2-4) # using the chain rule?

1 Answer
mason m
Feb 15, 2016

#f'(x)=-4xe^(-2x^2-4)#

Explanation:

The chain rule states that when differentiating a function that contains another function inside of it, the following happens: (1) differentiate the outside function and leave the inside function intact. (2) Multiply this by the derivative of the outside function.

The chain rule can be written as

#d/dx[g(h(x))]=g'(h(x))*h'(x)#

Here, we have #g(x)=e^x# and #h(x)=x^2-3x^2-4# which can be rewritten as #h(x)=-2x^2-4#.

Since the derivative of the outside function #g(x)=e^x# is still #g'(x)=e^x#, when we differentiate the outside function and leave the inner portion intact, we will still be left with

#overbrace(e^(-2x^2-4))^(g'(h(x)))#

Which is (the simplified version of) what we began with. Now, we just need to multiply this by the derivative of the inside function, #h(x)=-2x^2-4#. The power rule tells us that #h'(x)=-4x#.

Thus,

#f'(x)=overbrace(e^(-2x^2-4))^(g'(h(x)))*overbrace((-4x))^(h'(x))#

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