# How do you implicitly differentiate 3y + y^4/x^2 = 2?

Jul 7, 2016

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

#### Explanation:

${\left(3 y + {y}^{4} / {x}^{2} = 2\right)}^{p} r i m e$

[ie differentiate wrt x]

$\implies \left(3 y\right) ' + \left({y}^{4} / {x}^{2}\right) ' = {\left(2\right)}^{p} r i m e$

$\implies 3 y ' + \left({y}^{4} / {x}^{2}\right) ' = 0 q \quad \triangle$

two of those are very easy. we'll do the middle one using the quotient rule separately

$\left({y}^{4} / {x}^{2}\right) '$

$= \frac{{\left({y}^{4}\right)}^{p} r i m e {x}^{2} - {y}^{4} {\left({x}^{2}\right)}^{p} r i m e}{{x}^{4}}$

$= \frac{4 {y}^{3} y ' {x}^{2} - {y}^{4} \left(2 x\right)}{{x}^{4}}$ which we stuff straight back into $\triangle$ to get

$3 y ' + \frac{4 {y}^{3} y ' {x}^{2} - {y}^{4} \left(2 x\right)}{{x}^{4}} = 0$

some algebra

$3 y ' {x}^{4} + 4 {y}^{3} y ' {x}^{2} - 2 x {y}^{4} = 0$

$3 y ' {x}^{3} + 4 {y}^{3} y ' x - 2 {y}^{4} = 0$

collecting $y '$ s together

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

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