What is the derivative of #Tan^-1 (y/x)#?

3 Answers
Jul 6, 2018

Answer:

#d/dx(tan^-1(y/x))=y/(x-x^2sec^2y)#

Explanation:

#y=tan^-1(y/x)#
#tany=y/x#
Using the chain rule on the left side:
#d/dx(tany)=(sec^2y)y'#
Using the product rule on the right side:
#d/dx(y/x)=-y/x^2+(y')/x#
#:.(sec^2y)y'=-y/x^2+(y')/x#
#(sec^2y)y'-(y')/x=-y/x^2#
#y'(sec^2y-1/x)=-y/x^2#
#y'=-y/(x^2sec^2y-x)=y/(x-x^2sec^2y)#
#:.d/dx(tan^-1(y/x))=y/(x-x^2sec^2y)#

Answer:

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

Explanation:

#u=tan^-1(y/x)#

This problem needs a slight prerequisite of chain rule, and quotient rule.
#(du)/dx=1/(1+(y/x)^2)(y-xdy/dx)/x^2#

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

Thus,

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

Jul 6, 2018

It really depends upon what you are doing, and which independent variables matter to you.

Let # z(x,y) = tan^-1 (y/x) qquad qquad implies tan z = y/x qquad triangle#

Using the Quotient Rule and Implicit Differentiation on #triangle#:

# sec^2 z \ dz = ( x \ dy - y \ dx)/x^2 #

#:. dz = ( x^2)/(x^2+y^2) * \ ( x \ dy - y \ dx)/x^2 qquad square#

# implies dz = ( x \ dy - y \ dx) /(x^2+y^2) #

The partial derivatives are therefore:

  • #{(z_x = - y/(x^2 + y^2)),(z_y = x/(x^2 + y^2)):}#

But if #x# is the independent variable, ie #y = y(x)#, then you have from #square#:

  • # dz/dx = ( x \ dy/dx - y \ dx/dx) /(x^2+y^2) #

Which is the total derivative wrt #x#:

  • # dz/dx = z_y dy/dx + z_x#

#bb(implies d/dx (tan^-1 (y/x))= ( x \ y' - y ) /(x^2+y^2) ) #