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

Nov 8, 2016

$- 5 x - \frac{y}{x} = y '$

#### Explanation:

$5 {x}^{3} = - 3 x y + 2$ Differentiate the equation from the left and right

$15 {x}^{2} \left(1\right) = - 3 x y ' + - 3 y$ The product rule was used on -3xy

$15 {x}^{2} + 3 y = - 3 x y '$ Divide the -3x over to the other side

$- 5 x - \frac{y}{x} = y '$ The answer.

For -3xy the product rule states f g' + f' g
Say -3x is f and y is g
-3x y' + -3 y

When you differentiate, especially implicit differentiation, you follow what the rules say, and this one is saying $\frac{\mathrm{dy}}{\mathrm{dx}}$.
a $\frac{\mathrm{dx}}{\mathrm{dx}}$ causes a 1 (as seen when I did it on the $5 {x}^{3}$), and a $\frac{\mathrm{dy}}{\mathrm{dx}}$ will cause a y'. For other problems a $\frac{\mathrm{dx}}{\mathrm{dy}}$ will cause a x' and the $\frac{\mathrm{dy}}{\mathrm{dy}}$ will cause a 1.

Nov 8, 2016

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

#### Explanation:

$5 {x}^{3} = - 3 x y + 2$

Differentiate each term with respect to $x$

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

the first term on the RHS is differentiated using the product rule

$15 {x}^{2} = - 3 \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 0$

$15 {x}^{2} = - 3 y - 3 x \frac{\mathrm{dy}}{\mathrm{dx}}$

rearrange for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{15 {x}^{2} + 3 y}{- 3 x}$

cancel the 3 out.

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