Question

find the derivative of cosh(lnx)

Answer 1

\[\cosh(\ln(x))\] is a rational function

Answer 2

and that means...??

Answer 3

it is one polynomial over another

Answer 4

\[\cosh(x)=\frac{1}{2}(e^{-x}+e^x)\] replace \(x\) by \(\ln(x)\)

Answer 5

I thought it would be : \[\frac{ sech(lnx) }{ x }\]

Answer 6

you get
\[\frac{1}{2}(e^{-\ln(x)}+e^{\ln(x)})\]

Answer 7

which is \(\frac{1}{2x}+\frac{x}{2}\)

Answer 8

now it is easy to take the derivative

Answer 9

oohhhhhhhhhhhhhhh ok

Answer 10

good
easy now right?

Answer 11

wait, where did you get the 1/2x + x/2??

Answer 12

\[e^{\ln(x)}=x\]

Answer 13

wait, youre right! I just plugged in ln not ln(x) thanks!

Answer 14

yw

Answer 15

but where did the 1/2x come from? I understand the x/2...

Answer 16

\(e^{-\ln(x)}=\frac{1}{e^{\ln(x)}}=\frac{1}{x}\)