How do you find the derivative of the function $y = cos (\frac{1 - e^{2 x}}{1 + e^{2 x}})$ $y=cos((1-e^(2x))/(1+e^(2x)))$ ?

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Answer 1 · 2018-05-27T16:03:38

$y'=4*e^(2*x)/(1+e^(2+x))^2*sin((1-e^(2x))/(1+e^(2*x)))$

Using the chain rule, then we get
$y'=-sin((1-e^(2x))/(1+e^(2x)))*(-2e^(2x)(1+e^(2x))-(1-e^(2x))*e^(2x))/(1+e^(2x))^2$

$y=-sin((1-e^(2x))/(1+e^(2x)))*(2e^(2x)(-1-e^(2x)-1+e^(2x))/(1+e^(2x))^2)$
$y'=4e^(2x)/(1+e^(2x))^2*sin((1-e^(2x))/(1+e^(2x)))$

How do you find the derivative of the function y=cos((1-e^(2x))/(1+e^(2x)))y=cos(1−e2x1+e2x)y=cos((1-e^(2x))/(1+e^(2x)))?