# How do you find dy/dx by implicit differentiation of y^2=(x^2-4)/(x^2+4) and evaluate at point (2,0)?

Dec 6, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}}$ does not exist at $y = 0$. Finding #dy/dx takes a bit more work.

#### Explanation:

Beginning with ${y}^{2} = \frac{{x}^{2} - 4}{{x}^{2} + 4}$

we differentiate to get

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \text{some function of } x$.

But then $\frac{\mathrm{dy}}{\mathrm{dx}}$ requires dividing by $y$, so it is not defined at $y = 0$

In detail:

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

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{16 x}{{x}^{2} + 4} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 x}{y {\left({x}^{2} + 4\right)}^{2}}$

Here is the graph of the equation.

graph{y^2-(x^2-4)/(x^2+4)=0 [-8.077, 7.724, -4.394, 3.51]}