# How do you use implicit differentiation to find (d^2y)/(dx^2) given x^2-y^2=1?

Nov 2, 2016

$y ' ' \left(x\right) = \pm \frac{1}{{x}^{2} - 1} ^ \left(3 / 2\right)$

#### Explanation:

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

$\frac{\mathrm{df}}{\mathrm{dx}} = 2 x - 2 y \left(x\right) y ' \left(x\right) = 0$ and

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = 2 - 2 y \left(x\right) y ' ' \left(x\right) - 2 y ' {\left(x\right)}^{2} = 0$

so

$y ' ' \left(x\right) = \frac{y {\left(x\right)}^{2} - {x}^{2}}{y \left(x\right)} ^ 3 = \pm \frac{1}{{x}^{2} - 1} ^ \left(3 / 2\right)$