Derivatives of Trigonometric Functions with the Chain Rule

[physics604]

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.

 

[haruspex]

$=(1-e^{2x})(-(1+e^{2x})^{-2})+(1+e^{2x})^{-1}(-2e^{2x})$

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

$$= -\frac{1-e^{2x}}{(1+e^{2x})^2} + \frac{-2e^{2x}}{1+e^{2x}}$$
$$= \frac{-(1-e^{2x})}{(1+e^{2x})^2} + \frac{(-2e^{2x})(1+e^{2x})}{(1+e^{2x})^2}$$
$$= -\frac{-1+e^{2x}+(-2e^{2x})+(-2e^{4x})}{(1+e^{2x})^2}$$

Sign error.

 

[Student100]

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.

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.

 

[HallsofIvy]

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.

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

 

[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

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

 

[physics604]

Thanks everyone! I've got it!

 

[Student100]

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

Thanks?