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Thread: Check my derivatives solutions.

  1. #1
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    Check my derivatives solutions.

    Are my answers right?

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

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

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

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

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

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

    5. $\displaystyle h(x)=5x^2*LnX$
    My Answer:
    $\displaystyle h1(x)=10x*Lnx+5x^2*\frac{1}{x}$
    =$\displaystyle h1(x)=10x*Lnx+5x$
    Last edited by MathKidd; Mar 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: $\displaystyle y=(x^2-5)*e^x$

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

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

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

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


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

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