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Math Help - Differentiation of Inverse Trig Function

  1. #1
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    Post Differentiation of Inverse Trig Function

    Differention with repeat to x
    cot^-1(x)

    Answer: -2/(Squar root (1-9X^2)

    i stuck...
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  2. #2
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    Quote Originally Posted by sanikui View Post
    Differention with repeat to x
    cot^-1(x)

    Answer: -2/(Squar root (1-9X^2)

    i stuck...
    Use integration by parts: \int u \, dv = uv - \int v \, du.

    u = \cot^{-1} x \Rightarrow du = \, ....

    dv = dx \Rightarrow v = \, ....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Use integration by parts: \int u \, dv = uv - \int v \, du.

    u = \cot^{-1} x \Rightarrow du = \, ....

    dv = dx \Rightarrow v = \, ....
    My mistake. I misread differentiate as integrate.

    y = \cot^{-1} x \Rightarrow \cot y = x

    Now differentiate both sides wrt x (use the chain rule on the left hand side) and then make dy/dx the subject.
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  4. #4
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    ok i try.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    My mistake. I misread differentiate as integrate.

    y = \cot^{-1} x \Rightarrow \cot y = x

    Now differentiate both sides wrt x (use the chain rule on the left hand side) and then make dy/dx the subject.
    Spoilers in white:

    Using the chain rule: {\color{white} \frac{d \, \cot y}{dx} = \frac{d \, \cot y}{dy} \cdot \frac{dy}{dx} = -\frac{1}{\sin^2 y} \frac{dy}{dx}.}

    Note that: {\color{white} \cot y = x \Rightarrow \sin^2 y = \frac{1}{x^2 + 1}.}
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  6. #6
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    Finaly noe how to do this question. V^^
    Let x=coty

    x=1 over tany
    tany= 1 over x
    y= tan(1 over x)


    From there i can do liaw....thx alot ^^
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