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What is the derivative of the inverse tan(y/x)?

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May 24, 2015

The derivative would be #1/sqrt(x^2+y^2) (dy/dx -y/x)#

If u is #tan^-1(y/x)# then tan u =#y/x#. Differentiating w.r.t. x,

#sec^2u (du)/dx= 1/x^2 (xdy/dx -y)#

#(du)/dx= cos^2 u [1/x^2(x dy/dx -y)]#

= #x/sqrt(x^2+y^2)# #1/x^2 (xdy/dx -y)#

=#1/sqrt(x^2+y^2) (dy/dx -y/x)#

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