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

1 Answer
Feb 24, 2017

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

# y+y^3=x^2 #

Differentiate wrt #x#:

# \ \ \ \ 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 #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) = y+y^3-x^2 #; Then;

#(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.