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

1 Answer
Feb 24, 2017

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

Explanation:

When we differentiate #y# wrt #x# we get #dy/dx#.

However, we cannot differentiate a non implicit function of #y# wrt #x#. But if we apply the chain rule we can differentiate a function of #y# wrt #y# but we must also multiply the result by #dy/dx#.

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

We have:

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

Differentiate wrt #x# (applying product rule):

# (-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 #y# explicitly as a function of #x#, only implicitly through the equation #F(x, y) = 0# which defines #y# as a function of #x, y = y(x)#. Therefore we can write #F(x, y) = 0# as #F(x, y(x)) = 0#. Differentiating both sides of this, using the partial chain rule gives us

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

So Let # F(x,y) = 4x^3 + 2xy^2 + 3x^2y^3 - 3 #; Then;

#(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.