How do you differentiate # y =-2e^(xcosx)# using the chain rule?

1 Answer
Jun 6, 2017

Answer:

The answer is -2#e^(xcosx)##cosx-xsinx#.

Explanation:

The #d/dx# of #e^x# is always itself.

Now that we know this we move forward the trick here is that we take the #d/dx# of #e^x# then we apply the chain rule to #e^(xcosx)# we take the #d/dx# of the inside #xcosx#. Here we apply the product rule #f^1(g)# x #g^1(f)# the #d/dx# of this is #cosx-xsinx#. Our final answer is -2#e^(xcosx)##cosx-xsinx#. We don't take the 2 into account because its a constant.