Question

find teh derivative of
g(x) = cosh (lnx)

Answer 1

remember the definition of hyperbolic functions
\[Cosh(x)= \frac{ e^{x}-e^{-x} }{ 2 } \]
So,
\[Cosh[log(x)]= \frac{ e^{log(x)}-e^{-log(x)} }{ 2 } \;\;\; \rightarrow \; \;\; \;\; Cosh[log(x)]=\frac{x-(\frac{1}{x})}{2}\]
So,
\[(Cosh[log(x)]^{\prime}=\frac{1}{2}+\frac{2}{x^{2}} \]

Answer 2

y did u log everthing in teh 2nd step

Answer 3

I used the definition of Hyperbolic Cosine to simplify the expression, so instead of having x in the powers, I substituted them for "Logs" and simplified according to the rules of exponents.

Notice this is a composite function and the inside function is not "x" anymore but "log(x)."

Answer 4

There are other ways to do this this just a bit more transparent as to what's going on.

Answer 5

im confused with teh step where u have cosh (log x) = x - (1/x) / 2 and also with te hstep after it

Answer 6

I just realized there is typo....the sign in between the both "e" is a "+"...I was just typing without looking at the screen, sorry.......easy fix, at the that should be a "-."

Here, http://en.wikipedia.org/wiki/Hyperbolic_cosine scroll down a bit until you find Cosh(x)