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Math Help - find derivative

  1. #1
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    find derivative

    Find dy/dx if  y=(u^2+5)^{0.5}, u =(1-x)/(1+x); x=-2

    can someone check my work? let me know if i have anything wrong..

    y=({((1-x)/(1+x))^2+5})^{0.5}
    y=(1-x)/(1+x)+5^{0.5}
    dy/dx={(-1)(1+x)-(1-x)(1)}/(1+x)^2
    =(-2)/{(1+x)^2}
    at x=-2
    dy/dx=-2/{(1-2)^2}
    =-2
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    Find dy/dx if  y=(u^2+5)^{0.5}, u =(1-x)/(1+x); x=-2

    can someone check my work? let me know if i have anything wrong..

    y=({((1-x)/(1+x))^2+5})^{0.5}
    y=(1-x)/(1+x)+5^{0.5}
    dy/dx={(-1)(1+x)-(1-x)(1)}/(1+x)^2
    =(-2)/{(1+x)^2}
    at x=-2
    dy/dx=-2/{(1-2)^2}
    =-2
    I think you are supposed to do this using:

    \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    I think you are supposed to do this using:

    \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

    CB

    not sure what you mean by that, could you please elaborate and show me what you mean? Thank you.
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  4. #4
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    Hi skeske1234

    CaptainBlack told you to find \frac{dy}{du} and \frac{du}{dx} then multiply the results ^^


    EDIT : I checked your work and found this fatal mistake :
    y=({((1-x)/(1+x))^2+5})^{0.5}

    y=(1-x)/(1+x)+5^{0.5}

    Do you understand why this is wrong?
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  5. #5
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    Quote Originally Posted by skeske1234 View Post
    not sure what you mean by that, could you please elaborate and show me what you mean? Thank you.
    \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} is another representation of the chain rule.



    y = \sqrt{u^2 + 5}

    \frac{dy}{du} = \frac{u}{\sqrt{u^2+5}}

    u = \frac{1-x}{1+x}

    when x = -2 , u = -3 , \frac{dy}{du} = \frac{-3}{\sqrt{14}}


    \frac{du}{dx} = -\frac{2}{(1+x)^2}

    when x = -2 , \frac{du}{dx} = -2

    \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{-3}{\sqrt{14}} \cdot -2 = \frac{6}{\sqrt{14}}
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  6. #6
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    Quote Originally Posted by songoku View Post
    Hi skeske1234

    CaptainBlack told you to find \frac{dy}{du} and \frac{du}{dx} then multiply the results ^^


    EDIT : I checked your work and found this fatal mistake :
    y=({((1-x)/(1+x))^2+5})^{0.5}

    y=(1-x)/(1+x)+5^{0.5}

    Do you understand why this is wrong?
    Hi songoku, can you check my work below:

    dy/dx=(dy/du)(du/dx)

    =(u^2+5)^0.5 x -3
    =3(u^2+5)^0.5

    Question: why do I have to do this? i thought the question asked for the derivative of dy/dx and i substituted u into y and then found the derivative at x=-2.. isn't this what the question means?
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  7. #7
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    Quote Originally Posted by songoku View Post
    Hi skeske1234

    CaptainBlack told you to find \frac{dy}{du} and \frac{du}{dx} then multiply the results ^^


    EDIT : I checked your work and found this fatal mistake :
    y=({((1-x)/(1+x))^2+5})^{0.5}

    y=(1-x)/(1+x)+5^{0.5}

    Do you understand why this is wrong?
    ..... is it because im supposed to use the chain rule??
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  8. #8
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    Hi skeske1234
    Quote Originally Posted by skeske1234 View Post
    dy/dx=(dy/du)(du/dx)

    =(u^2+5)^0.5 x -3
    =3(u^2+5)^0.5
    I'm not sure what you are doing on this part...
    (u^2+5)^0.5 x -3 is not \frac{dy}{du} \cdot\frac{du}{dx}

    Question: why do I have to do this? i thought the question asked for the derivative of dy/dx and i substituted u into y and then found the derivative at x=-2.. isn't this what the question means?
    Yes, you can subs. u into y. But :

    <br />
(u^2+5)^{0.5} = \left(\left(\frac{1-x}{1+x}\right)^{2}+5\right)^{0.5} \neq \frac{1-x}{1+x} + 5^{0.5}<br />
    First of all, (a+b)^2 = a^2 + 2ab + b^2 not a^2+b^2

    Second, the formula above only applies for positive-integer power
    Last edited by songoku; August 10th 2009 at 10:25 AM.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by skeske1234 View Post
    not sure what you mean by that, could you please elaborate and show me what you mean? Thank you.
    First, the way the question is worded you are obviously supposed to use this relation (the chain rule)

    Second, the algebra is easier so you are less likely to make a mistake.

    CB
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