1. ## Hyperbolic functions

I am really not understanding these. For example i have

cosh6x+sinh6x

i know that cosh is 1/2(e^x + e^-x) and sinh is 1/2(e^x - e^-x)

so... what now?

1/2(e^x + e^-x)6x+1/2(e^x - e^-x)6x
??
i know somehow that 6x will be whats raised to it's power of the final answer which is e^6x but how?

2. Originally Posted by Frenchie
I am really not understanding these. For example i have

cosh6x+sinh6x

i know that cosh is 1/2(e^x + e^-x) and sinh is 1/2(e^x - e^-x)

so... what now?

1/2(e^x + e^-x)6x+1/2(e^x - e^-x)6x
??
i know somehow that 6x will be whats raised to it's power of the final answer which is e^6x but how?

$\cosh6x=\frac{e^{6x}+e^{-6x}}{2}$ , and similar for $\sinh6x$

Tonio

answer is -8tanh(8x+6) but i'm not sure how to go about solving it

4. First, do you really understand what your error in the first problem was? You wrote:

"cosh is 1/2(e^x + e^-x) and sinh is 1/2(e^x - e^-x)"

and then wrote

"cosh(6x)+ sinh(6x)= 1/2(e^x + e^-x)6x+1/2(e^x - e^-x)6x"

which is not only wrong it's "non-sense"- and it is important that you understand why. If that was not simply a typo or an oversight, then you have a serious misunderstanding of the notation.

"cosh", by itself, is NOT what you had. cosh(x)= (1/2)(e^x+ e^-x) and the "(x)" is crucial. cosh(6x) is NOT "cosh" multiplied by "6x", it is cosh(x) with x replaced by 6x:
cosh(6x)+ sinh(6x)= (1/2)(e^{6x}+ e^{-6x})+ (1/2)(e^{6x}- e^{-6x}).

Adding those, cosh(6x)+ sin(6x)= e^{6x} because the e^{6x} terms add and the e^{-6x} terms cancel.

answer is -8tanh(8x+6) but i'm not sure how to go about solving it"

The answer to what question? Don't you think that is an important piece of information?

I suspect you mean the answer to "the derivative of y with respect to x" but I don't know for sure.

What is the derivative of ln(x)? What is the derivative of $sech(x)= \frac{1}{cosh(x)}= \frac{2}{e^x+ e^{-x}}$?