# How do you implicitly differentiate (x/(x-4y))=x^(3)/y^2-1?

Jul 18, 2016

$y ' = \frac{2 {x}^{3} - {y}^{2} - 6 {x}^{2} y}{2 x y + 2 {x}^{3} - 6 {y}^{2}}$

#### Explanation:

$\left(\frac{x}{x - 4 y}\right) = {x}^{3} / {y}^{2} - 1 = \frac{{x}^{3} - {y}^{2}}{y} ^ 2$

$\therefore x {y}^{2} = \left(x - 4 y\right) \left({x}^{3} - {y}^{2}\right) = {x}^{4} - x {y}^{2} - 4 {x}^{3} y + 4 {y}^{3}$

$\therefore x {y}^{2} + x {y}^{2} + 4 {x}^{3} y = {x}^{4} + 4 {y}^{3}$

$\therefore 2 x {y}^{2} + 4 {x}^{3} y = {x}^{4} + 4 {y}^{3}$

Diff.ing both sides w.r.t. $x$,

$\frac{d}{\mathrm{dx}} \left(2 x {y}^{2}\right) + \frac{d}{\mathrm{dx}} \left(4 {x}^{3} y\right) = \frac{d}{\mathrm{dx}} \left({x}^{4}\right) + \frac{d}{\mathrm{dx}} \left(4 {y}^{3}\right)$

$\therefore 2 \left\{x \frac{d}{\mathrm{dx}} \left({y}^{2}\right) + {y}^{2} \frac{d}{\mathrm{dx}} \left(x\right)\right\} + 4 \left\{{x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {x}^{2} y\right\}$

$= 4 {x}^{3} + 4 \frac{d}{\mathrm{dx}} \left({y}^{3}\right) \ldots \ldots . . \left(1\right)$

Here, let us note that, by Chain Rule,

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

and on the same lines, $\frac{d}{\mathrm{dx}} \left({y}^{3}\right) = 3 {y}^{2} y '$.

These results will be used in $\left(1\right)$ [after cancelling $2$ throughout]

to get,

$\left\{x \cdot 2 y y ' + {y}^{2} \cdot 1\right\} + 2 \left\{{x}^{3} y ' + 3 {x}^{2} y\right\} = 2 {x}^{3} + 2 \cdot 3 {y}^{2} y '$

$2 x y y ' + {y}^{2} + 2 {x}^{3} y ' + 6 {x}^{2} y = 2 {x}^{3} + 6 {y}^{2} y '$

$y ' \left(2 x y + 2 {x}^{3} - 6 {y}^{2}\right) = 2 {x}^{3} - {y}^{2} - 6 {x}^{2} y$

$y ' = \frac{2 {x}^{3} - {y}^{2} - 6 {x}^{2} y}{2 x y + 2 {x}^{3} - 6 {y}^{2}}$