1. Calculus Theory:differentiating

Verify the identity:

arctanx+ arctan (1/x)=( ∏/2)

using calculus theory.

differentiate the left hand side of the identity.

2. Re: Calculus Theory:differentiating

If $f(x)=\arctan x+\arctan (1/x)\; (x\neq 0)$ then, $f'(x)=\ldots=0$ . As $(0,+\infty)$ is connected, then $f(x)$ is constant in $(0,+\infty)$. But $f(1)=\pi/4+\pi/4=\pi/2$ so $f(x)=\pi/2$ in $(0,+\infty)$ . Similar arguments for $(-\infty,0)$ .

3. Re: Calculus Theory:differentiating

I don't get it.

4. Re: Calculus Theory:differentiating

Originally Posted by LAPOSH42
I don't get it.
The derivative $f'(x)=\frac{1}{1+x^2}+\frac{-1}{x^2+1}=0$ for all $x\ne 0$.
So the function must be constant on its domain.

5. Re: Calculus Theory:differentiating

Originally Posted by Plato
The derivative $f'(x)=\frac{1}{1+x^2}+\frac{-1}{x^2+1}=0$ for all $x\ne 0$.
So the function must be constant on its domain.
Take as a counterexample:

$f(x)=\begin{Bmatrix} 1 & \mbox{ if }& x<0\\0 & \mbox{if}& x>0\end{matrix}$

Then, $f'(x)=0$ for all $x$ on its domain and $f(x)$ is not constant. As I said in answer #2 the domain must be connected, or equivalently (in $\mathbb{R}$) an interval.

6. Re: Calculus Theory:differentiating

Originally Posted by FernandoRevilla
Take as a counterexample:

$f(x)=\begin{Bmatrix} 1 & \mbox{ if }& x<0\\0 & \mbox{if}& x>0\end{matrix}$
Then, $f'(x)=0$ for all $x$ on its domain and $f(x)$ is not constant. As I said in answer #2 the domain must be connected, or equivalently (in $\mathbb{R}$) an interval.
Yes, I understand that. It was a quick reply to that particular problem. I should have mentioned the two different connected domain $(0,\infty)$.

If fact, the function is $\frac{-\pi}{2}$ on $(-\infty,0)$