# How do you implicitly differentiate  y^2/x= x^3 - 3yx^2 ?

Jul 10, 2016

Use the product and quotients rules and do a lot of tedious algebra to get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{4} + 2 {x}^{3} y + {y}^{2}}{2 x y + {x}^{4}}$.

#### Explanation:

We will begin on the left hand side:
${y}^{2} / x$

In order to take the derivative of this, we need to use the quotient rule:
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$

We have $u = {y}^{2} \to u ' = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$ and $v = x \to v ' = 1$, so:
$\frac{d}{\mathrm{dx}} \left({y}^{2} / x\right) = \frac{\left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(x\right) - \left({y}^{2}\right) \left(1\right)}{x} ^ 2$
$\to \frac{d}{\mathrm{dx}} \left({y}^{2} / x\right) = \frac{2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{2}}{x} ^ 2$

Now for the right hand side:
${x}^{3} - 3 y {x}^{2}$

We can use the sum rule and multiplication of a constant rule to break this into:
$\frac{d}{\mathrm{dx}} \left({x}^{3}\right) - 3 \frac{d}{\mathrm{dx}} \left(y {x}^{2}\right)$

The second of these will require the product rule:
$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$
With $u = y \to u ' = \frac{\mathrm{dy}}{\mathrm{dx}}$ and $v = {x}^{2} \to v ' = 2 x$. So:
$\frac{d}{\mathrm{dx}} \left({x}^{3} - 3 y {x}^{2}\right) = 3 {x}^{2} - \left(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left({x}^{2}\right) + \left(y\right) \left(2 x\right)\right)$
$\to \frac{d}{\mathrm{dx}} \left({x}^{3} - 3 y {x}^{2}\right) = 3 {x}^{2} - {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y$

Our problem now reads:
$\frac{2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{2}}{x} ^ 2 = 3 {x}^{2} - {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y$

We can add ${x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$ to both sides and factor out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ to isolate it:
$\frac{2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{2}}{x} ^ 2 = 3 {x}^{2} - {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y$
$\to \frac{2 x y \frac{\mathrm{dy}}{\mathrm{dx}}}{x} ^ 2 + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{{y}^{2}}{x} ^ 2 = 3 {x}^{2} + 2 x y$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{2 x y}{x} ^ 2 + {x}^{2}\right) = 3 {x}^{2} + 2 x y + \frac{{y}^{2}}{x} ^ 2$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 2 x y + \frac{{y}^{2}}{x} ^ 2}{\frac{2 x y}{x} ^ 2 + {x}^{2}}$

I hope you like algebra, because this is one nasty equation that needs to be simplified:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 2 x y + \frac{{y}^{2}}{x} ^ 2}{\frac{2 x y}{x} ^ 2 + {x}^{2}}$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{3 {x}^{4}}{x} ^ 2 + \frac{2 {x}^{3} y}{x} ^ 2 + \frac{{y}^{2}}{x} ^ 2}{\frac{2 x y}{x} ^ 2 + {x}^{4} / {x}^{2}}$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{3 {x}^{4} + 2 {x}^{3} y + {y}^{2}}{x} ^ 2}{\frac{2 x y + {x}^{4}}{x} ^ 2}$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{4} + 2 {x}^{3} y + {y}^{2}}{x} ^ 2 \cdot {x}^{2} / \left(2 x y + {x}^{4}\right)$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{4} + 2 {x}^{3} y + {y}^{2}}{2 x y + {x}^{4}}$