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Math Help - Inverse Cosine Derivative

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

    This is stumping me. Find the derivative of the following function:

    <br />
y = \cos ^{ - 1} \left( {\frac{x}<br />
{2}} \right)<br />

    We know this rule:

    <br />
\frac{d}<br />
{{dx}}\cos ^{ - 1} u = \frac{{ - du/dx}}<br />
{{\sqrt {1 - u^2 } }},|u| < 1<br />

    So.....


    <br />
\begin{gathered}<br />
  u = \tfrac{x}<br />
{2} \hfill \\<br />
  \tfrac{{du}}<br />
{{dx}} = \tfrac{1}<br />
{2} \hfill \\<br />
  \frac{d}<br />
{{dx}}\cos ^{ - 1} u = \frac{{ - 1/2}}<br />
{{\sqrt {1 - \tfrac{{x^2 }}<br />
{4}} }} = \frac{{ - 1/2}}<br />
{{\sqrt {4 - x^2 } }} \hfill \\ <br />
\end{gathered} <br />

    But this is wrong....it should be....

    <br />
y = \cos ^{ - 1} \left( {\frac{x}<br />
{2}} \right) = \frac{{ - 1}}<br />
{{\sqrt {4 - x^2 } }}<br />

    What's going on here?
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  2. #2
    Super Member 11rdc11's Avatar
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    \frac{{ - 1/2}}<br />
{{\sqrt {1 - \tfrac{{x^2 }}<br />
{4}} }}

    Multipy that by

    \frac{\sqrt{4}}{\sqrt{4}}
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  3. #3
    Member RedBarchetta's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    \frac{{ - 1/2}}<br />
{{\sqrt {1 - \tfrac{{x^2 }}<br />
{4}} }}

    Multipy that by

    \frac{\sqrt{4}}{\sqrt{4}}
    So this is true as well? Given a as a constant.

    <br />
\frac{d}<br />
{{dx}}\cos ^{ - 1} \frac{x}<br />
{a} = \frac{{ - 1}}<br />
{{\sqrt {a^2  - x^2 } }}<br />
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  4. #4
    Super Member 11rdc11's Avatar
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    I think so
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    So this is true as well? Given a as a constant.

    <br />
\frac{d}<br />
{{dx}}\cos ^{ - 1} \frac{x}<br />
{a} = \frac{{ - 1}}<br />
{{\sqrt {a^2  - x^2 } }}<br />

    Yup!

    Because \frac{d}{\,dx}\cos^{-1}\left(\frac{x}{a}\right)=-\frac{1}{a}\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}=-\frac{1}{\sqrt{a^2}}\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}=-\frac{1}{\sqrt{a^2\left(1-\frac{x^2}{a^2}\right)}}=-\frac{1}{\sqrt{a^2-x^2}}

    --Chris
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