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)#