Calculus Theory:differentiating

**LAPOSH42** · December 6th 2011, 09:22 PM

Verify the identity:

arctanx+ arctan (1/x)=( ∏/2)

using calculus theory.

differentiate the left hand side of the identity.

**FernandoRevilla** · December 6th 2011, 09:45 PM

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 $(-\infty,0)$ .

**LAPOSH42** · December 7th 2011, 12:33 PM

I don't get it.

**Plato** · December 7th 2011, 12:49 PM

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.

**FernandoRevilla** · December 7th 2011, 11:20 PM

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.

**Plato** · December 8th 2011, 02:41 AM

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 $(-\infty,0)$

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