What is the derivative of #Tan^-1 (y/x)#?
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#d/dx(tan^-1(y/x))=y/(x-x^2sec^2y)#
#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)#
#d/dx(tan^-1(y/x))=x^2(y-xdy/dx)/(x^2+y^2)#
#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)#
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) ) #
