# How do you differentiate (y^2+1)^2-x=0?

Apr 10, 2018

The implicit derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ is $\frac{1}{4 y \left({y}^{2} + 1\right)}$.

#### Explanation:

First, rearrange a little:

${\left({y}^{2} + 1\right)}^{2} - x = 0$

${\left({y}^{2} + 1\right)}^{2} = x$

Now, take the implicit derivative (which basically means treat $y$ as a function of $x$):

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

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

$2 {\left({y}^{2} + 1\right)}^{2 - 1} \cdot \frac{d}{\mathrm{dx}} \left[{y}^{2} - 1\right] = 1$

$2 {\left({y}^{2} + 1\right)}^{1} \cdot \frac{d}{\mathrm{dx}} \left[{y}^{2} - 1\right] = 1$

$2 \left({y}^{2} + 1\right) \cdot \frac{d}{\mathrm{dx}} \left[{y}^{2} - 1\right] = 1$

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{4 y \left({y}^{2} + 1\right)}$

That's the derivative. Hope this helped!