- #1
physics604
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Homework Statement
Find the derivative of $$y=cos(\frac{1-e^{2x}}{1+e^{2x}})$$
Homework Equations
Chain rule
The Attempt at a Solution
$$y=cosu$$ $$\frac{dy}{du}=-sinu$$
$$u=\frac{1-e^{2x}}{1+e^{2x}}$$ $$ \frac{du}{dx}=(1-e^{2x})(-(1+e^{2x})^{-2})+(1+e^{2x})^{-1}(-2e^{2x})$$ $$= -\frac{1-e^{2x}}{(1+e^{2x})^2} + \frac{-2e^{2x}}{1+e^{2x}}$$ $$= \frac{-(1-e^{2x})}{(1+e^{2x})^2} + \frac{(-2e^{2x})(1+e^{2x})}{1+e^(1+e^{2x})^2}$$ $$= -\frac{-1+e^{2x}+(-2e^{2x})+(-2e^{4x})}{(1+e^{2x})^2}$$
$$\frac{dy}{dx}=-sin\frac{1-e^{2x}}{1+e^{2x}} (-\frac{-1+e^{2x}+(-2e^{2x})+(-2e^{4x})}{(1+e^{2x})^2})$$
The front part of my answer is right $$-sin\frac{1-e^{2x}}{1+e^{2x}}$$, but the second half is wrong.
According to my textbook it is supposed to be $$\frac{4e^{2x}}{(1+e^{2x})^2}$$. What did I do wrong?
Any help is much appreciated.