How do you use implicit differentiation to find #(dy)/(dx)# given #sin2x^2y^3=3x^3+1#?

1 Answer
Jul 24, 2017

# dy/dx = (9x^2 - 4xcos(2x^2)y^3)/(3sin(2x^2)y^2) #

Explanation:

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

However, we only differentiate explicit functions of #y# wrt #x#. But if we apply the chain rule we can differentiate an implicit function of #y# wrt #y# but we must also multiply the result by #dy/dx#.

Example:

#d/dx(y^2) = d/dy(y^2)dy/dx = 2ydy/dx #

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

Now, we have:

# sin2x^2y^3=3x^3+1 #

Implicitly differentiating wrt #x# (applying product rule):

# (sin2x^2)(d/dx y^3) + (d/dx sin2x^2)(y^3) = 9x^2#

# :. (sin2x^2)(3y^2dy/dx ) + (4xcos2x^2)(y^3) = 9x^2#

# :. 3sin(2x^2)y^2dy/dx + 4xcos(2x^2)y^3 = 9x^2#

# :. 3sin(2x^2)y^2dy/dx = 9x^2 - 4xcos(2x^2)y^3 #

# :. dy/dx = (9x^2 - 4xcos(2x^2)y^3)/(3sin(2x^2)y^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) = sin2x^2y^3-3x^3-1 #; Then;

#(partial F)/(partial x) = 4xcos(2x^2)y^3-9x^2 #

#(partial F)/(partial y) = sin(2x^2)(3y^2) #

And so:

# dy/dx = -(4xcos(2x^2)y^3-9x^2)/(sin(2x^2)(3y^2)) #
# " " = (9x^2-4xcos(2x^2)y^3)/(3sin(2x^2)y^2) #, as before.