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Thread: Inverse Trig Derivative

  1. #1
    Member RedBarchetta's Avatar
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    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}
    $
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by RedBarchetta View Post
    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
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