How do you take the derivative of #Tan^-1 (y/x)#?

1 Answer
Jun 7, 2015

I'm assuming you are thinking of this as being a function of two independent variables #x# and #y#: #z=tan^{-1}(y/x)#. The answers are #\frac{\partial z}{\partial x}=-\frac{y}{x^{2}+y^{2}}# and #\frac{\partial z}{\partial y}=\frac{x}{x^2+y^2}#.

Both of these facts can be derived with the Chain Rule, the Power Rule, and the fact that #y/x=yx^{-1}# as follows:

#\frac{\partial z}{\partial x}=\frac{1}{1+(y/x)^2}\cdot \frac{\partial}{\partial x}(yx^{-1})=\frac{1}{1+(y/x)^2}\cdot (-yx^{-2})#

#=-\frac{y}{x^{2}+y^{2}}#

and

#\frac{\partial z}{\partial y}=\frac{1}{1+(y/x)^2}\cdot \frac{\partial}{\partial y}(yx^{-1})=\frac{1}{1+(y/x)^2}\cdot (x^{-1})#

#=\frac{1/x}{1+y^{2}/(x^{2})}=\frac{x}{x^2+y^2}#