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)))