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

2 Answers
May 3, 2016

Answer:

#-2(x+1)*e^(-x^2-2x+1)#

Explanation:

The chain rule basically says #dy/dx =(dy/dz)*(dz\/dx)# where # y=f(x)#

So in this case let #z=-x^2-2x+1#

Then #y = e^z#

#dz/dx = -2x-2#

#dy/dz = e^z#

#dy/dx = (-2x-2)*e^(-x^2-2x+1) = -2(x+1)*e^(-x^2-2x+1)#

May 3, 2016

Answer:

You can do it another way using the chain rule more indirectly, based on the properties of #e#.

#f'(x)=-2xe^(-x^2-2x+1)-2e^(-x^2-2x+1)#

Explanation:

The derivative of #e^x# is always the derivative of #x# multiplied by #e^x#, or

#d/dx e^x = x' * e^x#.

This is why the derivative of #e^x# is just #e^x# when #x# is a variable on its own rather than an embedded function, because the derivative of a single variable like #x# is just #1#, so

#d/dx e^x=1*e^x=e^x#.

However, when you have embedded functions, you have to do more complex differentials with #x#.

In the case above, you have #-x^2-2x+1# instead of #x#. Work out the derivative of this normally,

#d/dx (-x^2-2x+1)=-2x-2#

and multiply by #e# to its original power, to get the final answer of

#(-2x-2)e^(-x^2-2x+1)=-2xe^(-x^2-2x+1)-2e^(-x^2-2x+1)#