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Math Help - finding the derivative

  1. #1
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    finding the derivative

    Let .


    I got

    [(-28x^3)(sqrtx) - (7x^4)(1/2(sqrtx))] - [(3x^2)(sqrtx)+(x^3)(1/2sqrtx)]/[(x^3sqrtx)^2]


    what did i do wrong?
    thanks
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  2. #2
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    Re: finding the derivative

    Hello, habibixox!

    \text{Differentiate: }\:f(x) \;=\;-7x^4\sqrt{x} + \frac{7}{x^3\sqrt{x}}

    HEY! . . . Simplify first!

    f(x) \;=\;-7x^4\sqrt{x} + \frac{7}{x^3\sqrt{x}} \;=\;-7x^4x^{\frac{1}{2}} + \frac{7}{x^3x^{\frac{1}{2}}} \;=\;-7x^{\frac{9}{2}} + \frac{7}{x^{\frac{7}{2}}}

    . . f(x) \;=\;-7x^{\frac{9}{2}} + 7x^{-\frac{7}{2}}


    Therefore: . f'(x) \;=\;-\tfrac{63}{2}x^{\frac{7}{2}} - \tfrac{49}{2}x^{-\frac{9}{2}} \;=\;-\tfrac{7}{2}x^{-\frac{9}{2}}\left(9x^8 + 7\right)

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  3. #3
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    Re: finding the derivative

    wow that's so much neater xD

    thank you so much!!!
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  4. #4
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    Re: finding the derivative

    Quote Originally Posted by Soroban View Post
    Hello, habibixox!


    HEY! . . . Simplify first!

    f(x) \;=\;-7x^4\sqrt{x} + \frac{7}{x^3\sqrt{x}} \;=\;-7x^4x^{\frac{1}{2}} + \frac{7}{x^3x^{\frac{1}{2}}} \;=\;-7x^{\frac{9}{2}} + \frac{7}{x^{\frac{7}{2}}}

    . . f(x) \;=\;-7x^{\frac{9}{2}} + 7x^{-\frac{7}{2}}


    Therefore: . f'(x) \;=\;-\tfrac{63}{2}x^{\frac{7}{2}} - \tfrac{49}{2}x^{-\frac{9}{2}} \;=\;-\tfrac{7}{2}x^{-\frac{9}{2}}\left(9x^8 + 7\right)

    It's also common practice to write your answer in the same form as the question, i.e. since the question was given in square root form, the answer should be given in square root form...
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