# Derivatives, chain rule 4

1. Nov 9, 2013

### physics604

1. The problem statement, all variables and given/known data

Find the derivative of $$y=cos(\frac{1-e^{2x}}{1+e^{2x}})$$

2. Relevant equations

Chain rule

3. 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.

2. Nov 9, 2013

### haruspex

$=(1-e^{2x})(-(1+e^{2x})^{-2})$<something missing here>$+(1+e^{2x})^{-1}(-2e^{2x})$
Sign error.

3. Nov 9, 2013

### Student100

Why don't you just try using the quotient rule instead of the product rule? It's much cleaner, and using the product rule you made a mistake.

Edit:: Are you trying to wing yourself off using u sub for these yet? It helps, since I believe you forgot $(1+e^{2x})^{-1}$ is also a composition.

Last edited: Nov 9, 2013
4. Nov 10, 2013

### HallsofIvy

Staff Emeritus
"wing yourself off"? Oh, you mean "wean". That stopped me for a moment.

5. Nov 10, 2013

### lurflurf

There are several ways to approach these problems

I would be inclined to rewrite the function
$$\cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right)=\cos \left(1- \frac{2}{1+e^{2x}} \right)=\cos(\tanh(x))$$

Otherwise use the chain rule multiple times with care. For example

$$\cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right)=\mathrm{f}(\mathrm{g}(\mathrm{h}(x))) \\ \text{where} \\ \mathrm{f}(x)=\cos(x) \\ \mathrm{g}(x)=\frac{1-x}{1+x} \\ \mathrm{h}(x)=e^{2x}$$

6. Nov 10, 2013

### physics604

Thanks everyone! I've got it!

7. Nov 10, 2013

Thanks?