Question

What is the derivative of arctan(x)+arctan(1/x)?

Answer 1

`0`

Explanation:

Alternatively, we can simplify the original function.

`y=arctan(x)+arctan(1/x)`

Take the tangent of both sides.

`tan(y)=tan(arctan(x)+arctan(1/x))`

Use the tangent addition formula: `tan(alpha+beta)=(tan(alpha)+tan(beta))/(1-tan(alpha)tan(beta))`

Here, for `tan(arctan(x)+arctan(1/x))`, we see that `alpha=arctan(x)` and `beta=arctan(1/x)`, so we obtain:

`tan(y)=(tan(arctan(x))+tan(arctan(1/x)))/(1-tan(arctan(x))tan(arctan(1/x))`

`tan(y)=(x+1/x)/(1-x(1/x))`

`tan(y)=((x^2+1)/x)/(1-1)`

`tan(y)=(x^2+1)/0`

This is an undefined value: however, we know that the tangent of `y` is undefined, and the tangent function is undefined at `pi/2` and `-pi/2`.

So, we know that

`y=pi/2" "` or `" "y=-pi/2`

Thus,

`arctan(x)+arctan(1/x)=+-pi/2`

And the derivative of `pi/2` or `-pi/2`, which are both constant, is just `0`.