1. ## Derivative Help

Can somebody please give me the derivative of the following:

$\displaystyle A(x) = x^2 + \sqrt3(\frac{16-2x}{3})^2$

$\displaystyle \frac{16-2x}{3}$ is supposed to be inside brackets all squared. I don't know how to fix the brackets. Sorry.

2. Originally Posted by Morphayne
Can somebody please give me the derivative of the following:

$\displaystyle A(x) = x^2 + \sqrt3(\frac{16-2x}{3})^2$

$\displaystyle \frac{16-2x}{3}$ is supposed to be inside brackets all squared. I don't know how to fix the brackets. Sorry.
$\displaystyle A(x) = x^2 + \sqrt3(\frac{16-2x}{3})^2$

so

$\displaystyle 2x + 2\sqrt{3}(\frac{16-2x}{3})(\frac{-2}{3})$

3. Originally Posted by Morphayne
Can somebody please give me the derivative of the following:

$\displaystyle A(x) = x^2 + \sqrt3(\frac{16-2x}{3})^2$

$\displaystyle \frac{16-2x}{3}$ is supposed to be inside brackets all squared. I don't know how to fix the brackets. Sorry.

First, rewrite it like this:

$\displaystyle A(x) = x^2 + \frac{\sqrt{3}}{9} (16 - 2x)^2$

Then try it. Don't forget the chain rule for the $\displaystyle (16 - 2x)^2$

4. What do you guys get when you solve $\displaystyle A'(x)=0$?

5. Originally Posted by Morphayne
What do you guys get when you solve $\displaystyle A'(x)=0$?

I get:

$\displaystyle x = \frac{64}{6\sqrt{3} + 8}$