How do you differentiate #y=cos((1-e^(2x))/(1+e^(2x)))#?

1 Answer
Apr 7, 2018

#d/dxcos((1-e^(2x))/(1+e^(2x)))=(4e^(2x))/(1+e^(2x))^2sin((1-e^(2x))/(1+e^(2x)))#

Explanation:

The Chain Rule, when applied to the cosine, tells us that if #u# is some function in terms of #x,#

#d/dxcosu=-sin(u)*(du)/dx#

Here, we see #u=(1-e^(2x))/(1+e^(2x))#

#(du)/dx# can be calculated with the Quotient Rule and the Chain Rule for exponentials, which tells us that #d/dxe^(ax)# where #a# is some positive constant is given by #ae^(ax).#

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

Simplify.

#(du)/dx=(-2e^(2x)-2e^(4x)-2e^(2x)+2e^(4x))/(1+e^(2x))^2#

#(du)/dx=(-4e^(2x))/(1+e^(2x))^2#

Thus,

#d/dxcos((1-e^(2x))/(1+e^(2x)))=(-1)(-4)(e^(2x))/(1+e^(2x))^2sin((1-e^(2x))/(1+e^(2x)))#

#d/dxcos((1-e^(2x))/(1+e^(2x)))=(4e^(2x))/(1+e^(2x))^2sin((1-e^(2x))/(1+e^(2x)))#