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Math Help - Check my answers please

  1. #1
    Newbie
    Joined
    Feb 2009
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    23

    Check my answers please

    Differentiate

    1. f(x)= x^5(cosx)
    My answer: 5x^4(cosx) + x^5(-sinx)

    2. u= (2x+1)5^x
    My answer: 5^xIn5(2x+1) + (2)5^x

    3. g(s)= s^(5/3) - 2se^s
    My answer: 5s^(2/3)/3 - 2e^s + 2se^s

    4. y= 1/(x^3 + 3x + 1)
    My answer: 3x^2+3/(x^3 + 3x +1)^2

    5. y= 4/(e^t + 1)
    My answer: -4e^t/(e^t + 1)^2

    6. w= x/(x^2 + 2)
    My answer: -x^3 + 2x^2 -2x/ (x^2 + 2)^2
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    Quote Originally Posted by jkami View Post
    Differentiate

    1. f(x)= x^5(cosx)
    My answer: 5x^4(cosx) + x^5(-sinx)
    Correct.

    2. u= (2x+1)5^x
    My answer: 5^xIn5(2x+1) + (2)5^x
    Correct.

    3. g(s)= s^(5/3) - 2se^s
    My answer: 5s^(2/3)/3 - 2e^s + 2se^s
    Correct, only the final term must be -2se^s.

    4. y= 1/(x^3 + 3x + 1)
    My answer: 3x^2+3/(x^3 + 3x +1)^2
    Remember that \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}.

    5. y= 4/(e^t + 1)
    My answer: -4e^t/(e^t + 1)^2
    Correct.

    6. w= x/(x^2 + 2)
    My answer: -x^3 + 2x^2 -2x/ (x^2 + 2)^2
    I derived:

    <br />
\begin{aligned}<br />
\frac{dw}{dx}&=\frac{d}{dx}\left(\frac{x}{x^2+2}\r  ight)\\<br />
&=\frac{(x^2+2)\cdot 1 - x\cdot 2x}{(x^2+2)^2}\\<br />
&=\frac{x^2+2-2x^2}{(x^2+2)^2}\\<br />
&=\frac{-x^2+2}{(x^2+2)^2}.<br />
\end{aligned}<br />
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  3. #3
    Newbie
    Joined
    Feb 2009
    From
    Valparaíso, Chile
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    9
    You have an error when you differentiate the number 4, it must be:
    y'=\dfrac{-(3x^2+3)}{(x^3 + 3x +1)^2}.
    Also in the number 6:
    y'=\dfrac{\{x\}'(x^2+2)-x\{x^2+2\}'}{(x^2 + 2)^2}=\dfrac{x^2+2-2x^2}{(x^2 + 2)^2}=\dfrac{2-x^2}{(x^2+2)^2}
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Santiago, Chile
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    Another way to differentiate last question could be

    w(x)=\frac{x}{{{x}^{2}}+2}=\frac{1}{x+2{{x}^{-1}}}={{\left( x+2{{x}^{-1}} \right)}^{-1}}, then w'(x)=-\left( 1-\frac{2}{{{x}^{2}}} \right)\cdot \frac{1}{{{\left( x+2{{x}^{-1}} \right)}^{2}}}=\frac{2-{{x}^{2}}}{{{\left( {{x}^{2}}+2 \right)}^{2}}}.
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