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Math Help - Am I on the right track with quotient rule?

  1. #1
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    Am I on the right track with quotient rule?

    Differentiate with respect to x using the quotient rule?
    Y= (2x4 3x) / (4x -1)

    d/dx [2x4 3x] = 8x3 3
    and
    d/dx [4x ] = 4

    by the quotient rule

    f(x) = (g(x)) / (h(x)) , f(x) = [g1(x) h(x) g(x)h1(x)] / [h(x)]2

    choosing g(x) = 2x4 3x and h(x) = 4x - 1

    gives

    dy/dx = [(8x3 3) (4x 1) - (2x4 3x) (4) ] / (4x 1) 2
    therefore

    dy/dx = (23x4 8x3 12x + 3 8x4 + 12x) / 16x2 8x + 1

    therefore
    dy/dx = (24x4 8x3 + 3) / (16x2 8x + 1)
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  2. #2
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    Re: Am I on the right track with quotient rule?

    In the second last line, it is not clear where 23x^4 appeared from. Also, in the same line, division is done before subtraction, which is not what you want. The final result is correct.
    Thanks from tomjay
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  3. #3
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    Re: Am I on the right track with quotient rule?

    sorry a typo there was meant to be 32.

    could I check another one please

    Find the integral:

    ∫ (5x2 + √x- (4/x2)) dx

    = (5x3/3) + (2x3/2/3) + 4/x + c

    = ((5x3 + 2x3/2 ) / 3) + (4/x) + c , where c is a constant

    Or

    = ((5x4 + 2x5/2 + 12) / 3x )) + c , where c is a constant
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  4. #4
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    Re: Am I on the right track with quotient rule?

    Find the integral:
    ∫ (5x2 + √x- (4/x2)) dx
    = (5x3/3) + (2x 3/2/3) + 4/x + c
    = ((5x3 + 2x 3/2 ) / 3) + (4/x) + c , where c is a constant
    Or
    = ((5x4 + 2x 5/2 + 12) / 3x )) + c , where c is a constant
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  5. #5
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    Re: Am I on the right track with quotient rule?

    In plain text, x^y is written x^y. Also, it is important to write parentheses because 2x^{3/2} is completely different from \frac{2x^3}{2}. You could check the result yourself in WoldramAlpha. Just don't overuse it and do the work by hand first if you are supposed to.
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