Question

How do you find derivative of `f(x)=cosh(lnx)`?

Answer 1

`d/dxf(x) = 1/2(1 - 1/x^2)`

Explanation:

`f(y) = cosh y = (e^y + e^-y)/2`

Taking `y = ln x`
`cosh (ln x) = (x + 1/x)/2`
Taking the derivative with respect to x:
`d/dxf(x) = 1/2(1 - 1/x^2)`