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Math Help - finding the differiential of the function

  1. #1
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    finding the differiential of the function

    Can anyone teach me step by step how to differetiate?

    1) f(x)=2x^3+x

    2)f(x)=3/sqrt(x)

    3)f(x)=3x^5/6+7x^2/3
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    do you remember the power rule?

    \frac d{dx}x^n = nx^{n - 1}

    if there is a constant multiplying, it doesn't matter, just take it along

    \frac d{dx}cx^n = c \frac d{dx}x^n = cnx^{n - 1} for c a constant
    Quote Originally Posted by lemontea View Post
    Can anyone teach me step by step how to differetiate?

    1) f(x)=2x^3+x
    apply the power rule for each term

    2)f(x)=3/sqrt(x)
    note that this is the same as f(x) = 3x^{-1/2} and apply the power rule

    3)f(x)=3x^5/6+7x^2/3
    i am not sure what you wrote here. please use parentheses to clarify

    type fractions like (numerator)/(denominator)

    example, type (3x + 1)/(x - 2) to mean \frac {3x + 1}{x - 2}
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  3. #3
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    I'm assuming you dont need the proof of this from first principles,we would be using the rules that

    \frac{d}{dx}(ax^n)=anx^{n-1}

    and that

    \frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d  }{dx}(g(x))

    for the first one we would have in the first term the power (n) is 3 and the coefficient before it (a) is 2 so the differential of that would be 2 \times 3 x^2=6x^2

    then the second term a=1 and n=1 so the differential of that would be 1 giving us

    f'(x)=6x^2+1

    can you do the same for the other parts now in the second remember that \frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}
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