# How do you find dy/dx by implicit differentiation given y^2+3y-5=x?

Jan 25, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 y + 3}$

#### Explanation:

 y^2+3y−5=x

Differentiate wrt $x$ and we get:

 d/dx(y^2) + d/dx(3y) − d/dx(5) = d/dx (x)
 :. d/dx(y^2) + 3dy/dx − 0 = 1
$\therefore \frac{d}{\mathrm{dx}} \left({y}^{2}\right) + 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

This is where we get stuck because we can differentiate $y$ wrt $x$ to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ but we cannot different ${y}^{2}$ wrt $x$. We can however use the chain rule, to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{dy}} \left({y}^{2}\right) + 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

And we can now handle this:

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot 2 y + 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\therefore 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\therefore \left(2 y + 3\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 y + 3}$

we can skip the direct application of the chain rule and use the result it yields which is that if we differentiate a function of $y$ wrt $x$ then we get $\frac{\mathrm{dy}}{\mathrm{dx}}$ times the function of $y$ differentiated wrt $y$ - It is this that is known as "Implicit Differentiation".