How do you implicitly differentiate 4y^2= x^3y+y-x 4y2=x3y+yx?

1 Answer
Jan 8, 2016

To differentiate implicitly, simply differentiate each term with respect to xx, then rearrange for dy/dxdydx.

Answer:dy/dx=(3x^2y+1)/(8y-x^3-1)dydx=3x2y+18yx31

Explanation:

4y^2=x^3y+y-x4y2=x3y+yx
d/(dx)(4y^2)=d/(dx)(x^3y)+d/(dx)(y)-d/(dx)(x)ddx(4y2)=ddx(x3y)+ddx(y)ddx(x)

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

d/(dx)(4y^2)=8ydy/dxddx(4y2)=8ydydx

This is from applying the chain rule.

d/(dx)(x^3y)=x^3dy/dx+3x^2yddx(x3y)=x3dydx+3x2y

This is from applying the product rule.

d/(dx)(y)=dy/dxddx(y)=dydx

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

Putting that all together we get:

8ydy/dx=x^3dy/dx+3x^2y+dy/dx+18ydydx=x3dydx+3x2y+dydx+1

Rearrange:

dy/dx(8y-x^3-1)=3x^2y+1dydx(8yx31)=3x2y+1

dy/dx=(3x^2y+1)/(8y-x^3-1)dydx=3x2y+18yx31

Sometimes it is also possible to find yy in terms of xx in the original equation and sub back to get dy/dxdydx only in terms of xx.