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Math Help - simple derivative problem!

  1. #1
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    simple derivative problem!

    \sqrt{2x^2}
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  2. #2
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    Re: simple derivative problem!

    Root(2x^2) =(root2)(rootx^2)=(root2)x So derivative is root2
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  3. #3
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    Re: simple derivative problem!

    Quote Originally Posted by rabert1 View Post
    \sqrt{2x^2}
    1. This is a term and not a function, so ...

    2. Where are you stuck (and why?)?

    3. I assume that you want to differentiate f(x) = \sqrt{2x^2}

    If so: Re-write the equation of the function:

    f(x) = \sqrt{2x^2}~\implies~f(x) = \left \lbrace \begin{array}{rcl}x \cdot \sqrt{2}& if & x \ge 0 \\ -x \cdot \sqrt{2}& if & x < 0\end{array} \right.

    4. Continue!
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  4. #4
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    Re: simple derivative problem!

    This is where I am stuck. I have to find the equation of a line tangent to the curve f(x) = \frac{\sqrt{2x^3}}{2} at the point (2,2)
    but when I use the quotient rule to find the derivative I get stuck when differentiating \sqrt{2x^3}
    Why? Does the x^\frac{1}{2} (from simplifying the root) get distributed to 2 and x^3? If you know a simpler way of doing this please show me.
    Last edited by rabert1; April 17th 2012 at 11:33 PM.
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  5. #5
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    Re: simple derivative problem!

    Quote Originally Posted by rabert1 View Post
    This is where I am stuck. I have to find the equation of a line tangent to the curve f(x) = \frac{\sqrt{2x^3}}{2} at the point (2,2)
    but when I use the quotient rule to find the derivative I get stuck when differentiating \sqrt{2x^3}
    Why? Does the x^\frac{1}{2} (from simplifying the root) get distributed to 2 and x^3? If you know a simpler way of doing this please show me.
    I don't understand why you want to use the quotient rule because there isn't any variable in the numerator.

    Re-write the equation of the function:

    f(x) = \frac{\sqrt{2x^3}}{2} ~\implies~f(x)=\frac12 \cdot \sqrt{2} \cdot x^{\frac32}

    Now apply the power rule to differentiate this function.
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  6. #6
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    Re: simple derivative problem!

    Quotient rule not necessary here. f(x)=(root2)/2 times x^3/2

    When differentiating (root2)/2 stays in front and x^3/2 becomes (3/2)x^1/2 You want this when x=2 ( I get 3/2 )
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