# How do you find dy/dx by implicit differentiation of x^2-y^2=16?

##### 1 Answer
Nov 23, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$

#### Explanation:

If an equation does not express $y$ explicitly in terms of $x$ as in $y = f \left(x\right)$, and instead we have $g \left(y\right) = f \left(x\right)$, then when we differentiate we apply the chain rule so that we differentiate $g \left(y\right)$ wrt $y$ rather than $x$ as in:

$g \left(y\right) = f \left(x\right)$
$\therefore \frac{d}{\mathrm{dx}} g \left(y\right) = \frac{d}{\mathrm{dx}} f \left(x\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} g \left(y\right) = f ' \left(x\right)$
$\therefore g ' \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right)$

This will typically result in $\frac{\mathrm{dy}}{\mathrm{dx}}$ being a function of $x$ and $y$, rather than $\frac{\mathrm{dy}}{\mathrm{dx}}$ being a function of $x$ alone as we usually get

So for ${x}^{2} - {y}^{2} = 16$ we have:

$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) - \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(16\right)$
$\therefore 2 x - \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left({y}^{2}\right) = 0$
$\therefore 2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$