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Math Help - More Derivatives

  1. #1
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    More Derivatives

    Use the short cut rules, then simplify results I am going crazy please help me

    f(x)=1/x

    k(x)=2x/x^2+2

    v(x)=(x^2+3)^

    m(x)=-5(x^4+3)^8
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  2. #2
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    Quote Originally Posted by Hallah_az
    Use the short cut rules, then simplify results I am going crazy please help me

    f(x)=1/x
    k(x)=2x/x^2+2
    v(x)=(x^2+3)^
    m(x)=-5(x^4+3)^8
    Hello,

    to 1.:
    f(x)=x^{-1} \Longrightarrow \frac{df}{dx}=(-1) \cdot x^{-2}=- \frac{1}{x^2}

    to 2.: You have to use the quotient rule, if the function reads like this:
    k(x)=\frac{2x}{x^2+2}

    \frac{dk}{dx}=\frac{(x^2+2) \cdot 2-2x \cdot 2x}{(x^2+2)^2} =\frac{-2x^2+4}{(x^2+2)^2}

    to 3.: Your problem is not complete. So I'll use n as exponent. You have to plug in the right value. You have to use the chain rule:

    v(x)=(x^2+3)^n \longrightarrow \frac{dv}{dx}=n \cdot (x^2+3)^{n-1} \cdot 2x

    to 4.: You have to use the chain rule:

    m(x)=-5(x^4+3)^8 \Longrightarrow \frac{dm}{dx}=(-5) \cdot 8 \cdot (x^4+3)^7 \cdot 4x^3

    \frac{dm}{dx}=-160x^3(x^4+3)^7

    Greetings

    EB
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