How do you differentiate #(e^ (2x) - e^(-2x) ) ^ 2#?

1 Answer
Roy E.
Dec 9, 2016

#4(e^{4x}-e^{-4x})#

Explanation:

First, differentiate "something" squared as #2xx# something. Then multiply by whatever you get by differentiating the something:
#2(e^{2x}-e^{-2z})^1xx d/{dx}(e^{2x}-e^{-2x}) =2(e^{2x}-e^{-2z})xx(2e^{2x}+2e^{-2x})# (because #d/dx(e^{ax})=ae^{ax}#)
#=4(e^{2x}-e^{-2x})(e^{2x}+e^{-2x})#.
Remember #(a-b)(a+b)=(a^2-b^2)# and #e^{2x}xxe^{2x}=e^{4x}#

Related questions
See all questions in Chain Rule
Impact of this question
1604 views around the world
You can reuse this answer
Creative Commons License