# How do you find (dy)/(dx) given -4x^2y^3+2=5x^2+y^2?

##### 1 Answer
Oct 4, 2016

$y ' = \frac{8 x {y}^{3} + 10 x}{- 4 {x}^{2} - 3 {y}^{2} - 2 y}$

#### Explanation:

Remember that Implicit Differentiation is really just a special case of the Chain Rule.

Every time that we differentiate the a factor or term what includes the variable y we have to include a factor of $\frac{\mathrm{dy}}{\mathrm{dx}}$ or $y '$.

• For the first term, $- 4 {x}^{2} {y}^{3}$, we have to use the Product Rule and Power Rule .
• For the constant, $2$, we have to use the Constant Rule .
• For the term, $5 {x}^{2}$, use the Power Rule .
• For the term, ${y}^{2}$, use the Power Rule .

$- 4 {x}^{2} 3 {y}^{2} y ' + \left(- 8\right) x {y}^{3} + 0 = 10 x + 2 y y '$

Gather the terms with $y '$ on one side of the equations and other terms on the other side.

$- 4 {x}^{2} 3 {y}^{2} y ' - 2 y y ' = 8 x {y}^{3} + 10 x$

Factor out $y '$

$y ' \left(- 4 {x}^{2} 3 {y}^{2} - 2 y\right) = 8 x {y}^{3} + 10 x$

Isolate $y '$ by dividing both sides by $\left(- 4 {x}^{2} 3 {y}^{2} - 2 y\right)$

$y ' \frac{\cancel{- 4 {x}^{2} 3 {y}^{2} - 2 y}}{\cancel{- 4 {x}^{2} 3 {y}^{2} - 2 y}} = \frac{8 x {y}^{3} + 10 x}{- 4 {x}^{2} 3 {y}^{2} - 2 y}$

$y ' = \frac{8 x {y}^{3} + 10 x}{- 4 {x}^{2} 3 {y}^{2} - 2 y}$

I have a couple of tutorials on Implicit Differentiation here, https://www.youtube.com/playlist?list=PLsX0tNIJwRTxL9RSJY4wKpW1MFbQfA84w