How do you implicitly differentiate #4y^2= x^3y+y-x #?

1 Answer
Jan 8, 2016

To differentiate implicitly, simply differentiate each term with respect to #x#, then rearrange for #dy/dx#.

Answer:#dy/dx=(3x^2y+1)/(8y-x^3-1)#

Explanation:

#4y^2=x^3y+y-x#
#d/(dx)(4y^2)=d/(dx)(x^3y)+d/(dx)(y)-d/(dx)(x)#

I will take it piece by piece for the sake of clarity.

#d/(dx)(4y^2)=8ydy/dx#

This is from applying the chain rule.

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

This is from applying the product rule.

#d/(dx)(y)=dy/dx#

#d/(dx)(x)=1#

Putting that all together we get:

#8ydy/dx=x^3dy/dx+3x^2y+dy/dx+1#

Rearrange:

#dy/dx(8y-x^3-1)=3x^2y+1#

#dy/dx=(3x^2y+1)/(8y-x^3-1)#

Sometimes it is also possible to find #y# in terms of #x# in the original equation and sub back to get #dy/dx# only in terms of #x#.