# How do you use implicit differentiation to find (dy)/(dx) given 5y^2=2x^3-5y?

Aug 3, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {x}^{2}}{2 y + 1} .$

#### Explanation:

$5 {y}^{2} = 2 {x}^{3} - 5 y \Rightarrow 5 {y}^{2} + 5 y = 5 \left({y}^{2} + y\right) = 2 {x}^{3.}$

$\therefore \frac{d}{\mathrm{dx}} \left\{5 \left({y}^{2} + y\right)\right\} = \frac{d}{\mathrm{dx}} \left\{2 {x}^{3}\right\} .$

$\therefore 5 \frac{d}{\mathrm{dx}} \left\{{y}^{2} + y\right\} = 2 \frac{d}{\mathrm{dx}} \left\{{x}^{3}\right\} = 2 \left\{3 {x}^{2}\right\} = 6 {x}^{2.} . . \left(1\right) ,$ whereas,

using the Chain Rule,

$\frac{d}{\mathrm{dx}} \left\{{y}^{2} + y\right\} = \frac{d}{\mathrm{dx}} \left\{{y}^{2}\right\} + \frac{d}{\mathrm{dx}} \left\{y\right\} ,$

$= \frac{d}{\mathrm{dy}} \left\{{y}^{2}\right\} \cdot \frac{d}{\mathrm{dx}} \left\{y\right\} + \frac{\mathrm{dy}}{\mathrm{dx}} ,$

$= 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} .$

Utilising this in $\left(1\right) ,$ we get,

$5 \left\{2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}}\right\} = 5 \left(2 y + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} ,$ giving,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {x}^{2}}{5 \left(2 y + 1\right)}$.