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Math Help - A Few questions on differentiation

  1. #1
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    A Few questions on differentiation

    Hi
    I need help on the following:

    1)Differentiate with respect with x: sin^2(x)

    I got 2sin(x)cos(x), however i don't understand why the book's answer is sin(2x).

    2)Differentiate with respect with x: arcos(\sqrt{1-x^2})

    This is what i have done:

    \frac{dy}{dx} = \frac{-0.5(1-x^2)^{-0.5} * 2x}{\sqrt{1-(\sqrt{1-x^2})^2}}

    = \frac{x}{\sqrt{1-x^2}\sqrt{x^2}}


    What would i do next?

    3) Prove that \frac{d}{dx}(arccos(x) = \frac{-1}{\sqrt{1-x^2}}

    How would i approach this??

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following:

    1)Differentiate with respect with x: sin^2(x)

    I got 2sin(x)cos(x), however i don't understand why the book's answer is sin(2x).

    2)Differentiate with respect with x: arcos(\sqrt{1-x^2})

    This is what i have done:

    \frac{dy}{dx} = \frac{-0.5(1-x^2)^{-0.5} * 2x}{\sqrt{1-(\sqrt{1-x^2})^2}}

    = \frac{x}{\sqrt{1-x^2}\sqrt{x^2}}


    What would i do next?

    3) Prove that \frac{d}{dx}(arccos(x) = \frac{-1}{\sqrt{1-x^2}}

    How would i approach this??

    P.S
    sin(2x) = 2 sinx. cosx So your answer is correct!
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  3. #3
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    clarification

    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following:

    1)Differentiate with respect with x: sin^2(x)

    I got 2sin(x)cos(x), however i don't understand why the book's answer is sin(2x).

    2)Differentiate with respect with x: arcos(\sqrt{1-x^2})

    This is what i have done:

    \frac{dy}{dx} = \frac{-0.5(1-x^2)^{-0.5} * 2x}{\sqrt{1-(\sqrt{1-x^2})^2}}

    = \frac{x}{\sqrt{1-x^2}\sqrt{x^2}}


    What would i do next?

    3) Prove that \frac{d}{dx}(arccos(x) = \frac{-1}{\sqrt{1-x^2}}

    How would i approach this??

    P.S
    1.sin(2x)=sin(x+x)=sinxcosx+cosxsinx=2sinxcosx
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  4. #4
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    a better way

    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following:

    1)Differentiate with respect with x: sin^2(x)

    I got 2sin(x)cos(x), however i don't understand why the book's answer is sin(2x).

    2)Differentiate with respect with x: arcos(\sqrt{1-x^2})

    This is what i have done:

    \frac{dy}{dx} = \frac{-0.5(1-x^2)^{-0.5} * 2x}{\sqrt{1-(\sqrt{1-x^2})^2}}

    = \frac{x}{\sqrt{1-x^2}\sqrt{x^2}}


    What would i do next?

    3) Prove that \frac{d}{dx}(arccos(x) = \frac{-1}{\sqrt{1-x^2}}

    How would i approach this??

    P.S
    2.there is a better way to do the second question:let y=arc(cos(sqrt(1-x^2))).then cosy=sqrt(1-x^2)=>-sinydy/dx=((-1/2)2x)/sqrt(1-x^2).find out siny from the original expression. plug in the value.things should get cancelled out and you should get your answer.
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  5. #5
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    3. Just in case a picture helps...



    ... where



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    The general drift is...

    _________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  6. #6
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    i still don't understand how to do question 3. how would i start it?
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  7. #7
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    OK, in this version, you start with y = arccos x, top left.

    Substitute (or map to) cos theta for x, then differentiate with respect to theta using the chain rule. And this is equal to differentiating theta with respect to theta. So...

    \frac{dy}{d\theta}\ =\ \frac{dy}{dx} \frac{dx}{d\theta}\ =\ 1

    \Rightarrow\ \frac{dy}{dx}\ =\ \frac{1}{\frac{dx}{d\theta}}\ =\ \frac{-1}{\sin \theta}

    So, using Pythag,

    \Rightarrow\ \frac{dy}{dx}\ =\ \frac{-1}{\sqrt{1 - \cos^2 \theta}}

    Then back-substitute x for cos theta.

    A different way is to start with

    y = arccos x\ \Rightarrow\ x = \cos y

    Then differentiate with respect to x...



    Substitute arccos x for y...

    \Rightarrow\ \frac{dy}{dx}\ =\ \frac{-1}{\sin (\arccos x)}

    Then use a right-triangle diagram or look up at http://staff.jccc.net/swilson/trig/compositions.htm to simplify.
    Last edited by tom@ballooncalculus; April 21st 2010 at 02:57 AM.
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  8. #8
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    so basically i make x=cos\theta
    which then i substitute into original equation:

    = arccos(cos\theta)

    = \frac{-1}{\sqrt{1-cos^2(\theta)}}

    sub back into equation and i get this:
    =\frac{-1}{\sqrt{1-x^2}}
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  9. #9
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    Quote Originally Posted by Paymemoney View Post
    so basically i make x=cos\theta
    which then i substitute into original equation:

    arccos(cos\theta)

    \Rightarrow \frac{d}{dx} \arccos x = = \frac{-1}{\sqrt{1-cos^2(\theta)}}" alt="\Rightarrow \frac{d}{dx} \arccos x = = \frac{-1}{\sqrt{1-cos^2(\theta)}}" />

    sub back into equation and i get this:
    =\frac{-1}{\sqrt{1-x^2}}
    With slight correction, yes
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  10. #10
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    Quote Originally Posted by Pulock2009 View Post
    2.there is a better way to do the second question:let y=arc(cos(sqrt(1-x^2))).then cosy=sqrt(1-x^2)=>-sinydy/dx=((-1/2)2x)/sqrt(1-x^2).find out siny from the original expression. plug in the value.things should get cancelled out and you should get your answer.
    can you explain how you got this, i don't understand.
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  11. #11
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    Quote Originally Posted by Paymemoney View Post
    can you explain how you got this, i don't understand.
    I think (s)he is referring to what I described in my second post as 'a different way...'

    Edit:

    Maybe not, though...
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