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

3 Answers
Jul 6, 2018

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)

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) )