Calculus Theory:differentiating

• Dec 6th 2011, 09:22 PM
LAPOSH42
Calculus Theory:differentiating
Verify the identity:

arctanx+ arctan (1/x)=( ∏/2)

using calculus theory.

differentiate the left hand side of the identity.
• Dec 6th 2011, 09:45 PM
FernandoRevilla
Re: Calculus Theory:differentiating
If $\displaystyle f(x)=\arctan x+\arctan (1/x)\; (x\neq 0)$ then, $\displaystyle f'(x)=\ldots=0$ . As $\displaystyle (0,+\infty)$ is connected, then $\displaystyle f(x)$ is constant in $\displaystyle (0,+\infty)$. But $\displaystyle f(1)=\pi/4+\pi/4=\pi/2$ so $\displaystyle f(x)=\pi/2$ in $\displaystyle (0,+\infty)$ . Similar arguments for $\displaystyle (-\infty,0)$ .
• Dec 7th 2011, 12:33 PM
LAPOSH42
Re: Calculus Theory:differentiating
I don't get it.
• Dec 7th 2011, 12:49 PM
Plato
Re: Calculus Theory:differentiating
Quote:

Originally Posted by LAPOSH42
I don't get it.

The derivative $\displaystyle f'(x)=\frac{1}{1+x^2}+\frac{-1}{x^2+1}=0$ for all $\displaystyle x\ne 0$.
So the function must be constant on its domain.
• Dec 7th 2011, 11:20 PM
FernandoRevilla
Re: Calculus Theory:differentiating
Quote:

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

Take as a counterexample:

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

Then, $\displaystyle f'(x)=0$ for all $\displaystyle x$ on its domain and $\displaystyle f(x)$ is not constant. As I said in answer #2 the domain must be connected, or equivalently (in $\displaystyle \mathbb{R}$) an interval.
• Dec 8th 2011, 02:41 AM
Plato
Re: Calculus Theory:differentiating
Quote:

Originally Posted by FernandoRevilla
Take as a counterexample:

$\displaystyle f(x)=\begin{Bmatrix} 1 & \mbox{ if }& x<0\\0 & \mbox{if}& x>0\end{matrix}$
Then, $\displaystyle f'(x)=0$ for all $\displaystyle x$ on its domain and $\displaystyle f(x)$ is not constant. As I said in answer #2 the domain must be connected, or equivalently (in $\displaystyle \mathbb{R}$) an interval.

Yes, I understand that. It was a quick reply to that particular problem. I should have mentioned the two different connected domain $\displaystyle (0,\infty)$.

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