Question

How do you find the derivative of `cosh(ln x)`?

Answer 1

let `y = cosh(lnx)`
`=> y = 1/2*(e^(lnx) - e^(-lnx))`

`= 1/2*(e^(lnx) + e^(lnx^-1))`

`= 1/2(x + x^-1)`

`(dy)/(dx) = 1/2(1 + (-1)*x^-2) = 1/2((x^2 - 1)/x^2) = (x^2 - 1)/(2x^2)`