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

##### 1 Answer

# dy/dx = -(12x^2 + 2y^2 +6xy^3)/(4xy+9x^2y^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:

# -2xy^2 - 3x^2y^3 + 3 = 4x^3 #

Differentiate wrt

# (-2x)(2ydy/dx) + (-2)(y^2) + (- 3x^2)(3y^2dy/dx) + (- 6x)(y^3) = 12x^2 #

# :. -4xydy/dx -2y^2 - 9x^2y^2dy/dx - 6xy^3 = 12x^2 #

# :. -(4xy+9x^2y^2)dy/dx = 12x^2 + 6xy^3 + 2y^2 #

# :. dy/dx = -(12x^2 + 6xy^3 + 2y^2)/(4xy+9x^2y^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) = 12x^2 + 2y^2 +6xy^3#

#(partial F)/(partial y) = 4xy+9x^2y^2 #

And so:

# dy/dx = -(12x^2 + 2y^2 +6xy^3)/(4xy+9x^2y^2)# , as before.