Question

Would you differentiate this one differently than the last problem?

Answer 1

\[\LARGE f(x)=cosh(lnx)\]

Answer 2

nope same idea

Answer 3

sub then chain rule

Answer 4

I'd probably change it knowing\[\cosh(t)=\frac{ e^t+e^{-t} }{ 2 }\]

Answer 5

actually
take in exponential form

Answer 6

Honestly works either way I think

Answer 7

\[\cosh(lnx)=\frac{ e^(lnx)+e^-(lnx) }{ 2 }\]

Answer 8

e^lnx=x

Answer 9

e^-lnx=1/x

Answer 10

Yeah pretty much preference here.

Answer 11

the exponential is probably simpler

Answer 12

btw @g17 , if you use {} around your lnx it'll stay in the exponent

Answer 13

yeah i will do it next time

Answer 14

@Luigi0210 the chain rule is a general rule that I believe you can show to yourself just using the definition of a derivative with d/dx(f(g(x)))=f'(g(x))*g'(x)

Answer 15

Alright, just making sure, thanks!