How do you implicitly differentiate -1=y^3x-xy+x^4y 1=y3xxy+x4y?

1 Answer
Mar 23, 2016

dy/dx=(y-y^3-4yx^3)/(3xy^2-x+x^4)dydx=yy34yx33xy2x+x4

Explanation:

The equation is:

y^3x-xy+x^4y=-1y3xxy+x4y=1

Differentiating w.r.t. xx

We have to apply product rule:

[y^3*d/dxx+xd/dx*y^3]-[x*d/dxy+yd/dx*x]+[x^4*d/dx*y+y*d/dxx^4]=0[y3ddxx+xddxy3][xddxy+yddxx]+[x4ddxy+yddxx4]=0

[y^3+3xy^2dy/dx]-[x*dy/dx+y]+[x^4*dy/dx+4yx^3]=0[y3+3xy2dydx][xdydx+y]+[x4dydx+4yx3]=0

y^3+3xy^2dy/dx-xdy/dx-y+x^4dy/dx+4yx^3=0y3+3xy2dydxxdydxy+x4dydx+4yx3=0

3xy^2dy/dx-xdy/dx+x^4dy/dx=-y^3+y-4yx^33xy2dydxxdydx+x4dydx=y3+y4yx3

dy/dx[3xy^2-x+x^4]=y-y^3-4yx^3dydx[3xy2x+x4]=yy34yx3

dy/dx=(y-y^3-4yx^3)/(3xy^2-x+x^4)dydx=yy34yx33xy2x+x4