(Headbang)

Verify the identity:

arctanx+ arctan (1/x)=( ∏/2)

using calculus theory.

differentiate the left hand side of the identity.

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- Dec 6th 2011, 09:22 PMLAPOSH42Calculus Theory:differentiating
(Headbang)

Verify the identity:

arctanx+ arctan (1/x)=( ∏/2)

using calculus theory.

differentiate the left hand side of the identity. - Dec 6th 2011, 09:45 PMFernandoRevillaRe: 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 PMLAPOSH42Re: Calculus Theory:differentiating
I don't get it.

- Dec 7th 2011, 12:49 PMPlatoRe: Calculus Theory:differentiating
- Dec 7th 2011, 11:20 PMFernandoRevillaRe: Calculus Theory:differentiating
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 AMPlatoRe: Calculus Theory:differentiating