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Math Help - Inverse Trig Derivative

  1. #1
    Member RedBarchetta's Avatar
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    Inverse Trig Derivative

    Find y'.

    <br />
\begin{gathered}<br />
  \frac{d}<br />
{{dx}}\tan ^{ - 1} x = \frac{1}<br />
{{x^2  + 1}} \hfill \\<br />
  y = \tan ^{ - 1} \sqrt {\tfrac{{1 + x}}<br />
{{1 - x}}}  \hfill \\ <br />
\end{gathered} <br />

    So...Where did I go wrong?

    <br />
y' = \frac{1}<br />
{{\tfrac{{1 + x}}<br />
{{1 - x}} + 1}} = \frac{1}<br />
{{\tfrac{{1 + x}}<br />
{{1 - x}} + \tfrac{{1 - x}}<br />
{{1 - x}}}} = \frac{1}<br />
{{\tfrac{{1 + 1 + x - x}}<br />
{{1 - x}}}} = \frac{1}<br />
{{\tfrac{2}<br />
{{1 - x}}}} = \frac{{1 - x}}<br />
{2}<br />
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by RedBarchetta View Post
    Find y'.

    <br />
\begin{gathered}<br />
  \frac{d}<br />
{{dx}}\tan ^{ - 1} x = \frac{1}<br />
{{x^2  + 1}} \hfill \\<br />
  y = \tan ^{ - 1} \sqrt {\tfrac{{1 + x}}<br />
{{1 - x}}}  \hfill \\ <br />
\end{gathered} <br />

    So...Where did I go wrong?

    <br />
y' = \frac{1}<br />
{{\tfrac{{1 + x}}<br />
{{1 - x}} + 1}} = \frac{1}<br />
{{\tfrac{{1 + x}}<br />
{{1 - x}} + \tfrac{{1 - x}}<br />
{{1 - x}}}} = \frac{1}<br />
{{\tfrac{{1 + 1 + x - x}}<br />
{{1 - x}}}} = \frac{1}<br />
{{\tfrac{2}<br />
{{1 - x}}}} = \frac{{1 - x}}<br />
{2}<br />
    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
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