Question

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

Answer 1

`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 2

`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)`

Answer 3

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) )`