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Math Help - Derivative of this function

  1. #1
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    Derivative of this function

    How would I find the derivative of this function: f(x)=\sqrt{3x^3}?

    I'm only supposed to use the power rule and I'm having a little trouble.
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    f(x) = 3x^{3/2}

    now use the power rule
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    f(x) = 3x^{3/2}

    now use the power rule
    Thanks! Doing that I got (9\sqrt{x})/2 Is that correct?

    I entered the original function on wolfram alpha and got a different answer for the derivative:S
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    Its good!
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    Quote Originally Posted by pickslides View Post
    Its good!
    Thanks! Back to the actual question, however, I had to write an equation for the tangent of f(x) at (3,9)

    What I'd normally do is set f\prime(3) to find the slope of the tangent, which would equal to (9\sqrt{3})/2 and set that equal to (y-9)/(x-3) to find the equation in standard form.

    The back has the final answer as 9x-2y-9=0, can anyone tell me if I'm doing anything wrong or if its a mistake in the book? Thanks!
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  6. #6
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    I get \displaystyle y-f(3)= f'(3)(x-3)
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    Quote Originally Posted by youngb11 View Post
    How would I find the derivative of this function: f(x)=\sqrt{3x^3}?

    I'm only supposed to use the power rule and I'm having a little trouble.
    Quote Originally Posted by e^(i*pi) View Post
    f(x) = 3x^{3/2}

    now use the power rule
    I hope it hasn't escaped notice that \displaystyle \sqrt{3x^3} \neq 3x^{3/2}

    -Dan
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  8. #8
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    A slight correction

    Quote Originally Posted by youngb11 View Post
    How would I find the derivative of this function: f(x)=\sqrt{3x^3}?

    I'm only supposed to use the power rule and I'm having a little trouble.
    \sqrt{3x^3} = 3^{1/2} \cdot x^{3/2}. Now apply the power rule and check it out.
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