# What is the implicit derivative of 3=1/y -x^2 ?

Jan 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x {y}^{2}$ The trick is in finding the derivative of each term, for term containing $y$ find derivative as usual and put $\frac{\mathrm{dy}}{\mathrm{dx}}$ next to it to indicate it is differentiated with respect to $x$

#### Explanation:

$3 = \frac{1}{y} - {x}^{2}$

$3 = {y}^{-} 1 - {x}^{2}$

Differentiate both sides with respect to $x$

$\frac{d}{\mathrm{dx}} \left(3\right) = \frac{d}{\mathrm{dx}} \left({y}^{-} 1\right) - \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

$0 = - {y}^{- 1 - 1} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x$

$0 = - {y}^{-} 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x$

$0 = - \frac{1}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x$

Add 2x to both the sides

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

Multiply both sides by $- {y}^{2}$

$- 2 x {y}^{2} = \frac{\mathrm{dy}}{\mathrm{dx}}$

Answer $\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x {y}^{2}$