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I should evaluate in which areas/intervals this function is differentiable and then differentiate.

$$ \arctan\left({\sqrt{\frac{x+1}{x-1}}}\right) $$

So my approach would be: assume continuity and so differentiability of $ \arctan(x)$ and then check for undefined points in $ {\sqrt{\frac{x+1}{x-1}}} $ So its undefined whenever my numerator is zero, or the term inside the root is negative. I guess for my level at university this would be enough and I dont need to prove limits or something like that?

loop
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  • That certainly seems appropriate. What valid intervals for $x$ did you come up with? – abiessu Sep 19 '13 at 15:05
  • $ -1 \lt x \leq 1 $ is invalid. rest valid. – loop Sep 19 '13 at 15:07
  • That is a good start. Note that $\tan x$ covers the entire set of real numbers, so $\arctan x$ should invert the entire set of reals to an interval over which $\tan$ is valid. I agree with your valid interval assessment for $x$. – abiessu Sep 19 '13 at 15:11
  • $ \tan(x) = \frac{\sin(x)}{\cos(x)} $ so $ \tan(x) $ is undefined at $\pi/2$ but how does that help me? And how do I differentiate the partial functions .. for x >= 1 and x < 1? – loop Sep 19 '13 at 15:14
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    I meant to say that your range for $x$ is good, and I think it's okay to move on to differentiation. The differentiation step involves $f\circ g\circ {j(x)\over h(x)}$, are you already familiar with taking derivatives of composed functions and division of functions? For deriving over separated intervals, note the valid intervals and derive as usual. You won't have two different derivatives for the different ranges, so you only have to note where the derivative will not be valid. – abiessu Sep 19 '13 at 15:30
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    I dont get your f g j/x and stuff part. But I think i will need so study the derivative of arctan(u) - right? so I dont mind my invalid intervals at the derivation, I just note them on the paper? – loop Sep 19 '13 at 15:43

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Let $$ \begin{align*} u & :=\sqrt{\frac{x+1}{x-1}} \end{align*} $$ then $$ \begin{align*} \left(\arctan u\right)' & =\frac{u'}{1+u^{2}} \end{align*} $$

Recall derivative of $(\sqrt{y(x)})'=\frac{y'(x)}{2\sqrt{y(x)}}$ so derivative of $u$ :

\begin{align*} u' & =\frac{\frac{(x+1)'(x-1)-(x-1)'(x+1)}{(x-1)^{2}}}{2\sqrt{\frac{x+1}{x-1}}} \end{align*} $$ $$ \begin{align*} \left(\arctan u\right)' & =\frac{\frac{\frac{-2}{(x-1)^{2}}}{2\sqrt{\frac{x+1}{x-1}}}}{1+\frac{x+1}{x-1}} \end{align*}

Ömer
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