# How do you find dy/dx given x=y^2+1?

Feb 28, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \pm \frac{1}{2 \sqrt{x - 1}}$

#### Explanation:

Differentiating both sides of the equation with respect to $x$ we have:

$1 = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 y}$

Now, as $x = {y}^{2} + 1 > 1$ we have $y = \pm \sqrt{x - 1}$

and:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \pm \frac{1}{2 \sqrt{x - 1}}$

Feb 28, 2017

dy/dx = +-1/(2sqrt(x-1)

#### Explanation:

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

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

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$ [Implicit differentiation]

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 y}$

Since ${y}^{2} = x - 1$ then $y = \pm \sqrt{x - 1}$

:. dy/dx = +-1/(2sqrt(x-1)