1. ## Inverse Trig Derivative

Find y'.

$\displaystyle \begin{gathered} \frac{d} {{dx}}\tan ^{ - 1} x = \frac{1} {{x^2 + 1}} \hfill \\ y = \tan ^{ - 1} \sqrt {\tfrac{{1 + x}} {{1 - x}}} \hfill \\ \end{gathered}$

So...Where did I go wrong?

$\displaystyle y' = \frac{1} {{\tfrac{{1 + x}} {{1 - x}} + 1}} = \frac{1} {{\tfrac{{1 + x}} {{1 - x}} + \tfrac{{1 - x}} {{1 - x}}}} = \frac{1} {{\tfrac{{1 + 1 + x - x}} {{1 - x}}}} = \frac{1} {{\tfrac{2} {{1 - x}}}} = \frac{{1 - x}} {2}$

2. Hello,

Originally Posted by RedBarchetta
Find y'.

$\displaystyle \begin{gathered} \frac{d} {{dx}}\tan ^{ - 1} x = \frac{1} {{x^2 + 1}} \hfill \\ y = \tan ^{ - 1} \sqrt {\tfrac{{1 + x}} {{1 - x}}} \hfill \\ \end{gathered}$

So...Where did I go wrong?

$\displaystyle y' = \frac{1} {{\tfrac{{1 + x}} {{1 - x}} + 1}} = \frac{1} {{\tfrac{{1 + x}} {{1 - x}} + \tfrac{{1 - x}} {{1 - x}}}} = \frac{1} {{\tfrac{{1 + 1 + x - x}} {{1 - x}}}} = \frac{1} {{\tfrac{2} {{1 - x}}}} = \frac{{1 - x}} {2}$
You have to use the chain rule...

You can notice that there is a $\displaystyle dx$ in the first expression you wrote. Well, this is equivalent to the derivative of x. But because it's 1, you often forget it..

The chain rule states :

$\displaystyle (u(v(x)))'=v'(x) \cdot u'(v(x))$

Here, $\displaystyle v(x)=\sqrt{\frac{1+x}{1-x}}=\frac{\sqrt{1+x}}{\sqrt{1-x}} \implies v'(x)=\dots \text{ (quotient rule) }$

And $\displaystyle u(t)=\tan^{-1} t$, with $\displaystyle t=v(x)$

Good luck