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Math Help - Check my derivatives solutions.

  1. #1
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    Check my derivatives solutions.

    Are my answers right?

    1. Find second derivative: y=(x^2-5)*e^x

    My Answer:
    Find second derivative:
    y1=2x*e^x+(x^2-5)*e^x
    y2=(2x)*(e^x)+2(e^x)+(x^2-5)*(e^x)+(2x)*(e^x)

    Find Derivatives:
    1. f(x)=\frac{3-x^4}{x^2+5}
    My Answer:
    f1(x)=\frac{(-4x^3)(x^2+5)-(3-x^4)(2x)}{(x^2+5)^2}

    2. f(x)=(14-3x^2)^5
    My Answer:
    f1(x)= -6x*5u^4
    = f1(x)=-6x*5(14-3x^2)^4
    = f1(x)=-30x(14-3x^2)^4

    3. f(x)=\sqrt{x^2-3}
    My Answer;
    f(x)=(x^2-3)^\frac{1}{2}
    = f1(x)=2x*\frac{1}{2}(x^2-3)^\frac{-1}{2}
    = f1(x)=1.5(x^2-3)^\frac{-1}{2}

    4. g(x)=2Ln(x^5)
    My Answer:
    g1(x)=5x^4*\frac{2}{x^5} or (i think) g1(x)=5x^4*\frac{5x^4}{2ln}

    5. h(x)=5x^2*LnX
    My Answer:
    h1(x)=10x*Lnx+5x^2*\frac{1}{x}
    = h1(x)=10x*Lnx+5x
    Last edited by MathKidd; March 28th 2010 at 09:16 PM.
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  2. #2
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  3. #3
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    Quote Originally Posted by MathKidd View Post
    Are my answers right?

    1. Find second derivative: y=(x^2-5)*e^x

    My Answer:
    Find second derivative:
    y1=2x*e^x+(x^2-5)*e^x
    y2=(2x)*(e^x)+(e^x)+(x^2-5)*(e^x)+(2x)*(e^x)
    y''=(2x)*(e^x)+2(e^x)+(x^2-5)*(e^x)+(2x)*(e^x)

    Quote Originally Posted by MathKidd View Post
    Find Derivatives:
    1. f(x)=\frac{3-x^4}{x^2+5}
    My Answer:
    f1(x)=\frac{(-4x^3)(x^2+5)-(3-x^4)(2x)}{(x^2+5)^2}
    Looks good but can be simplified

    Quote Originally Posted by MathKidd View Post
    2. f(x)=(14-3x^2)^5
    My Answer:
    f1(x)= -6x*5u^4
    =f1(x)=-6x*5(14-3x^2)^4
    =f1(x)=-30x(14-3x^2)^4
    Looks good

    Quote Originally Posted by MathKidd View Post
    3. f(x)=\sqrt{x^2-3}
    My Answer;
    f(x)=(x^2-3)^\frac{1}{2}
    =f1(x)=2x*\frac{1}{2}(x^2-3)^\frac{-1}{2}
    =f1(x)=1.5(x^2-3)^\frac{-1}{2}
    f'(x)=x(x^2-3)^\frac{-1}{2}


    Quote Originally Posted by MathKidd View Post
    4. g(x)=2Ln(x^5)
    My Answer:
    g1(x)=5x^4*\frac{2}{x^5} or (i think) g1(x)=5x^4*\frac{5x^4}{2ln}
    g'(x)=5x^4*\frac{2}{x^5}=\frac{10}{x}

    Quote Originally Posted by MathKidd View Post
    5. h(x)=5x^2*LnX
    My Answer:
    h1(x)=10x*Lnx+5x^2*\frac{1}{x}
    =h1(x)=10x*Lnx+5x
    Looks good
    Last edited by ione; March 28th 2010 at 09:38 PM.
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