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    y=3(x-4)/2x^3
    y=x^3(3x+8/x) diff these i carry on getting the wrong answer
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    (Original post by Ganhad)
    y=3(x-4)/2x^3
    y=x^3(3x+8/x) diff these i carry on getting the wrong answer
    What answer do you get?

    For the first one,
    \displaystyle  y=\frac{3(x-4)}{2x^3}=\frac{3x-12}{2x^3}=\frac{3x}{2x^3}-\frac{12}{2x^3}=\frac{3}{2x^2}-\frac{6}{x^3}=\frac32x^{-2}-6x^{-3}

    Now differentiate it.

    For the second one,
     \displaystyle y=x^3\left(3x+\frac8{x}\right)=3  x^4 + 8x^2
    Differentiate it.
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    (Original post by TheStudentBoom)
    first one:

    y= 3x-12/ 2x^3
    y = 3x/2x^3 - 12/2x^3

    y = 3/2x^2 - 6/x^3 => y= (3/2 x^-2) - (6x^-3)
    dy/dx = -3x^-3 + 12x^-4

    therefore dy/dx = -3/x^3 + 12/x^4

    might have done something wrong there, but thats what I know. I'll do the second one in a seperate post.
    Your answer is wrong.

    By the way, please don't post full solutions. Full solutions are considered a last resort.
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    (Original post by raheem94)
    What answer do you get?

    For the first one,
    \displaystyle  y=\frac{3(x-4)}{2x^3}=\frac{3x-12}{2x^3}=\frac{3x}{2x^3}-\frac{12}{2x^3}=\frac{3}{2x^2}-\frac{6}{x^3}=\frac32x^2-6x^{-3}
    hey the last bit is 3/2 x^-2 - 6x^-3 right? not 3/2 x^2?
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    (Original post by raheem94)
    Your answer is wrong.

    By the way, please don't post full solutions. Full solutions are considered a last resort.
    Very sorry, I'll just delete my posts then.
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    (Original post by TheStudentBoom)
    hey the last bit is 3/2 x^-2 - 6x^-3 right? not 3/2 x^2?
    Thanks for indicating my mistake. I will edit my post.
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    (Original post by TheStudentBoom)
    Very sorry, I'll just delete my posts then.
    +rep to you
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    (Original post by raheem94)
    +rep to you
    book says something different i did get what you got -_- Damn Book could you show me this one aswell y=3sqrtx(x-2)^2
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    (Original post by Ganhad)
    book says something different i did get what you got -_- Damn Book could you show me this one aswell y=3sqrtx(x-2)^2
    What is the answer in the book for the previous question?
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    (Original post by Ganhad)
    book says something different i did get what you got -_- Damn Book could you show me this one aswell y=3sqrtx(x-2)^2
    Is this your new question  \displaystyle y=3\sqrt{x(x-2)^2}
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    no the root does not go over the bracket just over the x

    (Original post by raheem94)
    Is this your new question  \displaystyle y=3\sqrt{x(x-2)^2}
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    (Original post by Ganhad)
    no the root does not go over the bracket just over the x
     \displaystyle y=3\sqrt{x}(x-2)^2 = 3x^{\frac12}(x^2-4x+4)=3x^{\frac52}-12x^{\frac32}+12x^{\frac12}

    Now its simple to differentiate it.

    By the way, what is the answer of the previous question in the book?
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    im supposed to find the gradient of that curve where x = 3 the answer is supposed to be 29root3 could you check if you get that please
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    (Original post by raheem94)
     \displaystyle y=3\sqrt{x}(x-2)^2 = 3x^{\frac12}(x^2-4x+4)=3x^{\frac52}-12x^{\frac32}+12x^{\frac12}

    Now its simple to differentiate it.

    By the way, what is the answer of the previous question in the book?
    -3x^-3+6x^-4
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    (Original post by Ganhad)
    im supposed to find the gradient of that curve where x = 3 the answer is supposed to be 29root3 could you check if you get that please
    Which question is it?

    You have posted 3 questions in this thread, which is this?
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    (Original post by Ganhad)
    -3x^-3+6x^-4
    Is this the answer to \displaystyle  y=\frac{3(x-4)}{2x^3}=\frac{3x-12}{2x^3}=\frac{3x}{2x^3}-\frac{12}{2x^3}=\frac{3}{2x^2}-\frac{6}{x^3}=\frac32x^{-2}-6x^{-3} \ ?

    If this is the case then it is wrong, the correct answer should be  \displaystyle -3x^{-3}+18x^{-4}
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    i agree but could you see if u get 29 root 3 if you sub x=3 into   \displaystyle y=3\sqrt{x}(x-2)^2 = 3x^{\frac12}(x^2-4x+4)=3x^{\frac52}-12x^{\frac32}+12x^{\frac12}
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    (Original post by Ganhad)
    i agree but could you see if u get 29 root 3 if you sub x=3 into   \displaystyle y=3\sqrt{x}(x-2)^2 = 3x^{\frac12}(x^2-4x+4)=3x^{\frac52}-12x^{\frac32}+12x^{\frac12}
    Substituting x=3 in   \displaystyle y=3\sqrt{x}(x-2)^2 give  3\sqrt3

    Substituting x=3 in the derivative of   \displaystyle y=3\sqrt{x}(x-2)^2 gives  \displaystyle \frac{13\sqrt3}2
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    this book is wak -_- so many damn mistakes
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    (Original post by raheem94)
    Substituting x=3 in   \displaystyle y=3\sqrt{x}(x-2)^2 give  3\sqrt3

    Substituting x=3 in the derivative of   \displaystyle y=3\sqrt{x}(x-2)^2 gives  \displaystyle \frac{13\sqrt3}2
    the equation of a curve is y= x^2-5x-24/x

    find the gradient of the curve at each of the point were the curve crosses the x-axis how would you do this sorry for asking for so much help but im a thicko that has been doing this for neardy 1 1/2 days straight
 
 
 
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