can someone explain finding the derivative of
y=(cosh x)

Question

can someone explain finding the derivative of
 y=(cosh x)

anonymous · Answer

I converted it to (e^x+e^-x)/2 but looking at it, the derivative of cosh x is sinhx so then could i divide by cos to get ln by itself and have sinhx/coshx

anonymous · Answer

oh shoot its, y= ln(coshx)

anonymous · Answer

express cosh using exponentials!

anonymous · Answer

uhhh?

anonymous · Answer

$$\cosh x={1\over2}(e^{-x}+e^x)$$

anonymous · Answer

$$
y=\ln(e^{-x}+e^x)-\ln(2)\
y'=\ldots
$$

anonymous · Answer

get it?

anonymous · Answer

kind of, what do i do with the log then?

anonymous · Answer

you take the derivative..
$$y'={1\over e^{-x}+e^x}\cdot (-e^{-x}+e^x)=\tanh(x)$$
if by "log" you mean to base "e"

anonymous · Answer

isnt the derivative of ln 1/x?

anonymous · Answer

yes.. and then the chain rule

anonymous · Answer

and it is not $\ln(x)$ but it is $\ln(e^{-x}+e^x)$

anonymous · Answer

ohhh. pellet. thanks.

anonymous · Answer

you are welcome.

anonymous · Answer

couldnt you also just keep it in sinhx and coshx form? seems simple that way.

anonymous · Answer

up to you..
I, being a human try to remember as much less as possible prioritizing the basics.
I know that cosh can be expanded and hence had no trouble..
there might be standard cosh derivative but cannot take chances.
this way is foolproof for me

anonymous · Answer

Alright thanks. i like knowing both ways :)