What is the derivative of tan^-1 (xy) = 1+ x^2y?

1 Answer
Apr 17, 2016

dy/dx=(2xy+2x^3y^3-y)/(x-x^2-x^4y^2)

Explanation:

We will be differentiating implicitly. On the left hand side, we will use the chain rule in regards to the inverse tangent function:

d/dx(arctan(u))=(u')/(1+u^2)

Also, note that the product rule will be used for d/dx(x^2y) and, eventually, d/dx(xy).

Differentiating gives:

(d/dx(xy))/(1+(xy)^2)=yd/dx(x^2)+x^2d/dx(y)

(yd/dx(x)+xd/dx(y))/(1+x^2y^2)=2xy+x^2dy/dx

(y+xdy/dx)/(1+x^2y^2)=2xy+x^2dy/dx

From here, just algebra your way through to an equation solved for dy/dx.

y+xdy/dx=(2xy+x^2dy/dx)(1+x^2y^2)

y+xdy/dx=2xy(1+x^2y^2)+dy/dx(x^2+x^4y^2)

xdy/dx-dy/dx(x^2+x^4y^2)=2xy+2x^3y^3-y

dy/dx(x-x^2-x^4y^2)=2xy+2x^3y^3-y

dy/dx=(2xy+2x^3y^3-y)/(x-x^2-x^4y^2)