# How do you find the derivative of inverse trig functions y= arctan(x^2-1)^(1/2) + arc csc(x) when x>1?

Oct 9, 2015

${y}^{'} = 0$

#### Explanation:

Assuming the equation was meant to be read as

$y = \arctan \left(\sqrt{{x}^{2} - 1}\right) + a r c \csc \left(x\right)$

We can say the derivative will be the sum of the two other derivatives

$u = \arctan \left(\sqrt{{x}^{2} - 1}\right)$
$v = a r c \csc \left(x\right)$

$v$ will be easier to differentiate, know, knowing that

$\csc \left(a r c \csc \left(x\right)\right) = x$

We differentiate both sides and use the chain rule, so

$1 = {\csc}^{'} \left(a r c \csc \left(x\right)\right) \cdot a r c {\csc}^{'} \left(x\right)$
$\frac{1}{\csc} ^ ' \left(a r c \csc \left(x\right)\right) = a r c {\csc}^{'} \left(x\right)$

We know that the derivative of the cossecant is $- \csc \left(x\right) \cot \left(x\right)$, so, in other words

$\frac{1}{- \csc \left(a r c \csc \left(x\right)\right) \cot \left(a r c \csc \left(x\right)\right)} = a r c {\csc}^{'} \left(x\right)$

Since they're both rational functions we can bring them up to the numerator

$- \tan \frac{a r c \csc \left(x\right)}{x} = a r c {\csc}^{'} \left(x\right)$

Using the pythagorean identity $1 + {\cot}^{2} \left(\theta\right) = {\csc}^{2} \left(\theta\right)$ we have

$1 + \frac{1}{{\tan}^{2} \left(a r c \csc \left(x\right)\right)} = {x}^{2}$

$\frac{1}{{\tan}^{2} \left(a r c \csc \left(x\right)\right)} = {x}^{2} - 1$

$\tan \left(a r c \csc \left(x\right)\right) = \frac{1}{\sqrt{{x}^{2} - 1}}$ so

$a r c {\csc}^{'} \left(x\right) = - \frac{1}{x \sqrt{{x}^{2} - 1}}$

(There are authors that say it like this, there are people that have that leading x as $| x |$, considering we're working with $x > 1$, the absolute value bars aren't needed regardless of your philosophy if they should be there or not).

Now, we do $u = \arctan \left(\sqrt{{x}^{2} - 1}\right)$, for the sake of brevity, I'll skip the proof of the arctan's derivative, as the process is much like the one used for the arccsc, except it'd involve the ${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$ identity.

So we have

$u = \arctan \left(\sqrt{{x}^{2} - 1}\right)$
${u}^{'} = {\arctan}^{'} \left(\sqrt{{x}^{2} - 1}\right) \cdot {\sqrt{\left({x}^{2} - 1\right)}}^{'} \cdot 2 x$

It's important to remember we need to apply the chain rule twice here.

${u}^{'} = \frac{1}{1 + {\left(\sqrt{{x}^{2} - 1}\right)}^{2}} \cdot \frac{1}{2 \sqrt{{x}^{2} - 1}} \cdot 2 x$
${u}^{'} = \frac{1}{1 + {x}^{2} - 1} \cdot \frac{1}{2 \sqrt{{x}^{2} - 1}} \cdot 2 x$
${u}^{'} = \frac{1}{x} ^ 2 \cdot \frac{1}{2 \sqrt{{x}^{2} - 1}} \cdot 2 x$
${u}^{'} = \frac{1}{x \sqrt{{x}^{2} - 1}}$

Summing them up we have

${y}^{'} = \frac{1}{x \sqrt{{x}^{2} - 1}} - \frac{1}{x \sqrt{{x}^{2} - 1}} = 0$

Which makes sense if you look at the graph for $x > 0$. For smaller values the absolute value probably would have mattered and would have made the derivative non-zero.