Hyperbolic Differentiation

1. Mar 30, 2014

Gondur

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

y=1 / (cos h x), find dy/dx

2. Relevant equations

chain rule and coshx=(e^x+e^-x)/2

3. The attempt at a solution

2. Mar 30, 2014

Curious3141

Resize your image!!! Way too big.

The most obvious error is $\frac{dy}{du} = \ln |u|$. Sure that's the derivative, and not the integral?

Other than that, for Chain Rule, it's generally not helpful to substitute variables like this. Not wrong, but it can overcomplicate things.

3. Mar 30, 2014

Gondur

What an idiot I am. This mistake is proof that I am tired and should get some sleep. Well what other method do you suggest I use, I'd definitely like to know if it's more efficient.

You can resize my image by clicking Ctrl and - on the keyboard. If you have a mouse with a wheel then turn the wheel towards you (downwards) while holding down Ctrl

4. Mar 30, 2014

SteamKing

Staff Emeritus
Start with the definition of cosh(x) in terms of the exponentials and re-write 1/cosh(x). What is dy/dx when y = 1/x or x^-1?

5. Mar 30, 2014

Gondur

6. Mar 30, 2014

Curious3141

Essentially what you did, but you don't have to spell out every step explicitly.

You're using $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{da}\frac{da}{dx}$, but there's no need to spell out every step.

Are you allowed to assume the derivative of the standard hyperbolic functions, like $\frac{d}{dx}\cosh x = \sinh x$?

If so, isn't it easier to let $u = \cosh x$ and $y = \frac{1}{u}$? Again, you shouldn't need to spell it out like this in different variables, this is just for clarity.

Yes, I know how to resize images - which is how I viewed yours. The problem is that it becomes impossible to view your text (and mine, when I post) in the same sizing. The font becomes too small.

If you're going to post here regularly, you would be much better served by learning basic LaTex.