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.