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Math Help - Differentiate questions

  1. #1
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    Differentiate questions

    Hi

    I need help on the following two questions:

    1)Differentiate with respect with x: \frac{1}{(7x^2+3)^5}

    This is what i have done, not sure if i have done it correctly:

    using quotient rule

    \frac{(7x^2+3)^5 * 0- 1 * 5(7x^2+3)^4 * 14x}{((7x^2+3)^5)^2}

    \frac{-70x(7x^2+3)^4}{(7x^2+3)^6}

    \frac{-70x}{(7x^2+3)^6}

    2)Differentiate with respect with x: (2x-1)^4 * (x+5)^{\frac{1}{2}}

    using the product rule

    (2x-1)^4 * \frac{1}{2}(x+5)^{\frac{1}{2}} + (x+5)^{\frac{1}{2}} * 4(2x-1)^3 * 2

    \frac{2(2x-1)^4}{\sqrt{x+5}} + \frac{8(2x-1)^3 * (x+5)^{\frac{1}{2}}}{\sqrt{x+5}}

    What would i do next?

    P.S
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  2. #2
    Junior Member piglet's Avatar
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    Quote Originally Posted by Paymemoney View Post
    Hi

    I need help on the following two questions:

    1)Differentiate with respect with x: \frac{1}{(7x^2+3)^5}

    This is what i have done, not sure if i have done it correctly:

    using quotient rule

    \frac{(7x^2+3)^5 * 0- 1 * 5(7x^2+3)^4 * 14x}{((7x^2+3)^5)^2}

    \frac{-70x(7x^2+3)^4}{(7x^2+3)^6}

    \frac{-70x}{(7x^2+3)^6}

    2)Differentiate with respect with x: (2x-1)^4 * (x+5)^{\frac{1}{2}}

    using the product rule

    (2x-1)^4 * \frac{1}{2}(x+5)^{\frac{1}{2}} + (x+5)^{\frac{1}{2}} * 4(2x-1)^3 * 2

    \frac{2(2x-1)^4}{\sqrt{x+5}} + \frac{8(2x-1)^3 * (x+5)^{\frac{1}{2}}}{\sqrt{x+5}}

    What would i do next?

    P.S
    Well for question 1, just bring the denominator up to the top line where it becomes  (7x^{2}+3)^{-5} . Then just use the chain rule... However your solution using the quotient rule is correct

    On question 2, your forgeting to reduce the power by 1 when differentiating
     ( x + 5 )^{\frac{1}{2}} ...

    Are you familar with how the chain rule works?
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  3. #3
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    ok i redone it and this is what i got:

    \frac{(2x-1)^4}{2\sqrt{x+5}} + \frac{16(x+5)(2x-1)^3}{2\sqrt{x+5}}

    Would i expand it??
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  4. #4
    Junior Member piglet's Avatar
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    well when i worked it out i got

     \frac{dy}{dx} = \frac{(2x-1)^{4}}{2\sqrt{x+5}} + 8(2x - 1)^3.\sqrt{x+5}

    There is no need to expand that but i guess you could tidy it up a little by using  2\sqrt{x+5} as your common denominator...
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  5. #5
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    Well how did the book's answer get this:

    \frac{(2x-1)^3(18x+79)}{2\sqrt{x+5}}
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  6. #6
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    Quote Originally Posted by Paymemoney View Post
    Well how did the book's answer get this:

    \frac{(2x-1)^3(18x+79)}{2\sqrt{x+5}}
    Read the last sentence of post #4 again.
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  7. #7
    Junior Member piglet's Avatar
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    Quote Originally Posted by Paymemoney View Post
    Well how did the book's answer get this:

    \frac{(2x-1)^3(18x+79)}{2\sqrt{x+5}}
    Ok i go through this with you...

    Using the common denominator  2\sqrt{x+5}

    you get  \frac{dy}{dx} = \frac{(2x-1)^{4} + 2\sqrt{x+5}(8(2x-1)^{3}.\sqrt{x+5})}{2\sqrt{x+5}}

    Tidying up this you get

     \frac{dy}{dx} = \frac{(2x-1)^{4} + 16(2x-1)^{3}.(x+5)}{2\sqrt{x+5}} as  \sqrt{x+5}.\sqrt{x+5} = x+5

    then taking out  (2x-1)^{3} as a common factor of the top line

    you get  \frac{dy}{dx} = \frac{(2x-1)^3(2x-1+16x+80)}{2\sqrt{x+5}}

    so tidying up what's in the brackets leaves you with

     \frac{dy}{dx} = \frac{(2x-1)^{3}(18x+79)}{2\sqrt{x+5}}

    Hope that answers your question..

    Piglet
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  8. #8
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    thank you i understand now.
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