# How do you find the derivative of the inverse of y = x^2 + 1?

y'=1/(2y)=1/(+-2sqrt(x-1)

#### Explanation:

The given function is

$y = {x}^{2} + 1$

The inverse is

$x = {y}^{2} + 1$

Differentiate both sides of the equation with respect to x

$\frac{d}{\mathrm{dx}} \left(x\right) = \frac{d}{\mathrm{dx}} \left({y}^{2} + 1\right)$

$1 = 2 y y ' + 0$

$y ' = \frac{1}{2 y}$

but $y = \pm \sqrt{x - 1}$

therefore

y'=1/(2*(+-sqrt(x-1)))=1/(+-2*sqrt(x-1)

God bless....I hope the explanation is useful.