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

##### 1 Answer

# dy/dx = (2x)/(1+3y^2) #

#### Explanation:

When we differentiate

However, we cannot differentiate a non implicit function of

When this is done in situ it is known as implicit differentiation.

We have:

# y+y^3=x^2 #

Differentiate wrt

# \ \ \ \ dy/dx+3y^2dy/dx = 2x #

# :. dy/dx(1+3y^2) = 2x #

# :. dy/dx = (2x)/(1+3y^2) #

**Advanced Calculus**

There is another (often faster) approach using partial derivatives. Suppose we cannot find

# (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)) #

So Let

#(partial F)/(partial x)=-2x \ \ \ # and#\ \ \ (partial F)/(partial y) = 1+3y^2 #

And so:

# dy/dx = -(-2x) / (1+3y^2) = (2x) / (1+3y^2) # , as before.