Question

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

Answer 1

`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)`