$\displaystyle \sqrt{2x^2}$

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- Apr 17th 2012, 11:11 PMrabert1simple derivative problem!
$\displaystyle \sqrt{2x^2}$

- Apr 17th 2012, 11:21 PMbiffboyRe: simple derivative problem!
Root(2x^2) =(root2)(rootx^2)=(root2)x So derivative is root2

- Apr 17th 2012, 11:22 PMearbothRe: simple derivative problem!
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 $\displaystyle f(x) = \sqrt{2x^2}$

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

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

4. Continue! - Apr 17th 2012, 11:31 PMrabert1Re: simple derivative problem!
This is where I am stuck. I have to find the equation of a line tangent to the curve $\displaystyle 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 $\displaystyle \sqrt{2x^3}$

Why? Does the $\displaystyle x^\frac{1}{2}$ (from simplifying the root) get distributed to $\displaystyle 2$ and $\displaystyle x^3$? If you know a simpler way of doing this please show me. - Apr 17th 2012, 11:54 PMearbothRe: simple derivative problem!
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:

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

Now apply the power rule to differentiate this function. - Apr 18th 2012, 12:01 AMbiffboyRe: 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 )