# Math Help - derivative of the inverse of a function

1. ## derivative of the inverse of a function

the function is y=x+(1/x) at y=17/4

attempt:
y^-1: x=y+(1/y)

differentiate: 1=y'+ln(y)y'
1=y'(1+ln(y))
y'=1/(1+ln(y))

put that over 1: 1+ln(y)

plug in y: 1+ln(17/4)
=approximately 2.447

2. ## Re: derivative of the inverse of a function

$\text{If}~ y(x)=x+\frac{1}{x} ~\text{then}~x(y)=\frac{1}{2} \cdot (y \pm \sqrt {y^2-4})$