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Thread: differentiate and simplify #1

  1. August 9th 2009, 03:28 PM #1
    skeske1234
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    differentiate and simplify #1

    y={(7x-2)^{0.5}/(5x+3)}^{0.5}

    can someone check my work? thanks

    y'=0.5{(7x-2)/(5x+3)}^{-0.5}{(7)(5x+3)-(7x-2)(5)}/{(5x+3)^2}
    y'=0.5{(7x-2)/(5x+3)}^{-0.5}{(31)/{(5x+3)^2}}
    y'=(31)/{2(5x+3)^{1.5}(7x-2)^{0.5}}
    Last edited by mr fantastic; August 12th 2009 at 03:56 PM. Reason: Changed post title
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  2. August 9th 2009, 03:45 PM #2
    skeeter
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    Quote Originally Posted by skeske1234 View Post
    y={(7x-2)/(5x+3)}^{0.5}

    can someone check my work? thanks

    y'=0.5{(7x-2)/(5x+3)}^(-0.5){(7)(5x+3)-(7x-2)(5)}/{(5x+3)^2}
    y'=0.5{(7x-2)/(5x+3)}^(-0.5){(31)/{(5x+3)^2}}
    y'=(31)/{2(5x+3)^{1.5}(7x-2)^{0.5}}

    y = \frac{7x-2}{\sqrt{5x+3}}<br />

    using the quotient rule ...

    y' = \frac{\sqrt{5x+3} \cdot 7 - (7x-2) \cdot \frac{5}{2\sqrt{5x+3}}}{5x+3}

    multiply numerator and denominator by 2\sqrt{5x+3} ...

    y' = \frac{14(5x+3) - 5(7x-2)}{2(5x+3)^{\frac{3}{2}}}

    y' = \frac{35x + 52}{2(5x+3)^{\frac{3}{2}}}
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  3. August 9th 2009, 03:47 PM #3
    skeske1234
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    Quote Originally Posted by skeeter View Post
    y = \frac{7x-2}{\sqrt{5x+3}}<br />

    using the quotient rule ...

    y' = \frac{\sqrt{5x+3} \cdot 7 - (7x-2) \cdot \frac{5}{2\sqrt{5x+3}}}{5x+3}

    multiply numerator and denominator by 2\sqrt{5x+3} ...

    y' = \frac{14(5x+3) - 5(7x-2)}{2(5x+3)^{\frac{3}{2}}}

    y' = \frac{35x + 52}{2(5x+3)^{\frac{3}{2}}}

    oops.. so sorry! the question was actually
    y={(7x-2)^{0.5}/(5x+3)}^{0.5}
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  4. August 9th 2009, 03:49 PM #4
    pickslides
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    I'm not sure what method you have used.

    I would apply the quoient rule which says y= \frac{u}{v} \Rightarrow y' =\frac{vu'-uv'}{v^2}

    In your case u = (7x-2) and v = (5x+3)^{0.5}

    Now find u' and v' and you should be fine.
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  5. August 9th 2009, 03:51 PM #5
    tom@ballooncalculus
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    Quote Originally Posted by skeske1234 View Post
    oops.. so sorry! the question was actually
    y={(7x-2)^{0.5}/(5x+3)}^{0.5}
    ... and your working is fine, wobbly latex aside

    Edit: wow, pardon me for mis-approving the original post - lots of editing was going on... well, no - no excuse.
    Last edited by tom@ballooncalculus; August 9th 2009 at 10:45 PM.
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  6. August 9th 2009, 03:59 PM #6
    skeeter
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    Quote Originally Posted by skeske1234 View Post
    oops.. so sorry! the question was actually
    y={(7x-2)^{0.5}/(5x+3)}^{0.5}
    you're not making this easy.


    y = \left(\frac{7x-2}{5x+3}\right)^{\frac{1}{2}}

    y' = \frac{1}{2}\left(\frac{7x-2}{5x+3}\right)^{-\frac{1}{2}} \cdot \left(\frac{(5x+3)(7) - (7x-2)(5)}{(5x+3)^2}\right)<br />

    y' = \frac{1}{2}\left(\frac{5x+3}{7x-2}\right)^{\frac{1}{2}} \cdot \left(\frac{31}{(5x+3)^2}\right)<br />

    y' = \frac{31}{2(7x-2)^{\frac{1}{2}}(5x+3)^{\frac{3}{2}}}<br />
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  7. August 9th 2009, 05:14 PM #7
    tom@ballooncalculus
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    Just in case a different (and I think comparably simple) route is of interest...



    No quotient rule as such; we tweak a few balloons (on the bottom row) ...



    ... and we have the final denominator ready - just simplify the numerator.


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