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Math Help - Differentiate with respect to x

  1. #1
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    Differentiate with respect to x

    Hi everyone, which way should I approach this one....

    Differentiate with respect to x

    Y = 3 arccot 2
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  2. #2
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    Quote Originally Posted by HNCMATHS View Post
    Hi everyone, which way should I approach this one....

    Differentiate with respect to x

    Y = 3 arccot 2
    since 3arccot(2) is a constant, the derivative is 0.

    if you meant y = 3arccot(2x) , then ...

    \dfrac{d}{dx} [k \cdot arccot(u)]  = -\dfrac{k}{1 + u^2} \cdot \dfrac{du}{dx}
    Last edited by skeeter; January 30th 2011 at 04:13 AM. Reason: had arctan on the brain ...
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  3. #3
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    Quote Originally Posted by skeeter View Post
    since 3\arctan(2) is a constant, the derivative is 0.

    if you meant y = 3\arctan(2x) , then ...

    \dfrac{d}{dx} [k\arctan{u}]  = \dfrac{k}{1 + u^2} \cdot \dfrac{du}{dx}
    Hi thanks for the reply, however the problem I am working with is definately 3arccot 2t
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    Quote Originally Posted by HNCMATHS View Post
    Hi everyone, which way should I approach this one....

    Differentiate with respect to x

    Y = 3 arccot 2
    Well, you can either just accept that

    \displaystyle\frac{d}{dx}\left[\cot^{-1}(x)\right]=\frac{-1}{1+x^2}

    or

    If y=\cot^{-1}(x), then x=\cot(y)

    Now derive x=\cot(y).

    \displaystyle 1=-\csc^2(y)\frac{dy}{dx}

    \displaystyle\frac{dy}{dx}=-\frac{1}{\csc^2(y)}

    \csc=\sqrt{1+\cot^2}=\sqrt{1+x^2}

    \displaystyle\frac{dy}{dx}=-\frac{1}{1+x^2}
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  5. #5
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    Quote Originally Posted by HNCMATHS View Post
    Hi everyone, which way should I approach this one....

    Differentiate with respect to x

    Y = 3 arccot 2
    I suppose you mean y = 3 arcot(2t) ....? (Which is exactly the point skeeter made in post #2 that you seem not to have grasped).
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    Quote Originally Posted by HNCMATHS View Post
    Hi everyone, which way should I approach this one....

    Differentiate with respect to x

    Y = 3 arccot (2x)
    arccot(x)=cot^{-1}(x)

    \displaystyle\frac{d}{dx}cot^{-1}(x)=-\frac{1}{1+x^2}

    \displaystyle\frac{d}{dx}cot^{-1}(u)=-\frac{1}{1+u^2}\;\frac{du}{dx} from the chain rule

    which gives

    \displaystyle\frac{d}{dx}cot^{-1}(2x)=?

    Using the notation "t"

    \displaystyle\frac{d}{dt}cot^{-1}(2t)=-\frac{1}{1+(2t)^2}\;\frac{d}{dt}(2t)
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