# Inverse Trig Derivative

• May 24th 2008, 11:22 AM
RedBarchetta
Inverse Trig Derivative
Find y'.

$
\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?

$
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}
$
• May 24th 2008, 11:31 AM
Moo
Hello,

Quote:

Originally Posted by RedBarchetta
Find y'.

$
\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?

$
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 $dx$ 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 :

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

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

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

Good luck :D