simple derivative problem!

**rabert1** · April 17th 2012, 11:11 PM

$\sqrt{2x^2}$

**biffboy** · April 17th 2012, 11:21 PM

Root(2x^2) =(root2)(rootx^2)=(root2)x So derivative is root2

**earboth** · April 17th 2012, 11:22 PM

Originally Posted by rabert1

$\sqrt{2x^2}$

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!

**rabert1** · April 17th 2012, 11:31 PM

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.

**earboth** · April 17th 2012, 11:54 PM

Originally Posted by rabert1

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.

**biffboy** · April 18th 2012, 12:01 AM

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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