Question

what is the derivative of this function ? 2x*sqrt(4-x)

Answer 1

Start with product rule:
\[\Large\sf \left(2x \sqrt{4-x}\right)'\quad=\quad \color{royalblue}{(2x)'}\sqrt{4-x}+2x\color{royalblue}{(\sqrt{4-x})'}\]There is the setup, need to take the derivative of the blue parts.

Answer 2

This is a good derivative to memorize, it shows up a lot,\[\Large\sf \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}\]

Answer 3

i got stuck after taking the lcd

Answer 4

Oh you wanted to combine the fractions to simplify?

Answer 5

yes

Answer 6

\[\Large\sf \left(2x \sqrt{4-x}\right)'\quad=\quad \color{royalblue}{(2x)'}\sqrt{4-x}+2x\color{royalblue}{(\sqrt{4-x})'}\]\[\Large\sf \left(2x \sqrt{4-x}\right)'\quad=\quad \color{orangered}{(2)}\sqrt{4-x}+2x\color{royalblue}{(\sqrt{4-x})'}\]\[\Large\sf \left(2x \sqrt{4-x}\right)'\quad=\quad \color{orangered}{(2)}\sqrt{4-x}+2x\cdot\color{orangered}{\frac{-1}{2\sqrt{4-x}}}\]Did you remember the negative that shows up from the chain rule?

Answer 7

yes i did

Answer 8

Cancel the twos and we're left to combine these two terms,\[\Large\sf \sqrt{4-x}-\frac{x}{\sqrt{4-x}}\]So it looks like the first term is missing a sqrt(4-x) in the denominator, yes? :o

Answer 9

Ah I missed a two on the first term :d my bad

Answer 10

once you put 2 on the first term is there anyway to simplify that more?

Answer 11

\[\frac{ 2\sqrt{4-x} }{ \sqrt{4-x} }-\frac{ x }{ \sqrt{4-x}}\]

Answer 12

First term needs a sqrt(4-x) for common denominator.
So multiply the first term by sqrt(4-x) top and bottom:\[\Large\sf \color{green}{\left(\frac{\sqrt{4-x}}{\sqrt{4-x}}\right)}\cdot2\sqrt{4-x}-\frac{x}{\sqrt{4-x}}\]Yah you have the right idea, but you forgot to give the top a sqrt(4-x) also.

Answer 13

do the sqrt cancel on the first term ?

Answer 14

on the numerator i mean

Answer 15

\[\Large\sf \color{green}{\left(\frac{\sqrt{4-x}}{\sqrt{4-x}}\right)}\cdot2\sqrt{4-x}-\frac{x}{\sqrt{4-x}}\]\[\Large\sf =\frac{2\sqrt{4-x}^2}{\sqrt{4-x}}-\frac{x}{\sqrt{4-x}}\]Cancel? Ehhh I guess if you want to think of it that way.
You have a sqrt being squared away

Answer 16

\[\Large\sf \frac{2(4-x)}{\sqrt{4-x}}-\frac{x}{\sqrt{4-x}}\]

Answer 17

to get the critical number, i have to set this equal to zero right?

Answer 18

Ya after you combine the fractions,
\[\Large\sf 0=\frac{8-3x}{\sqrt{4-x}}\]Setting equal to zero will give us critical numbers.

Answer 19

so x = 8/3?

Answer 20

@zepdrix = \(\LaTeX\) master!

Answer 21

Ok good.
I guess we would also want to include values which are undefined in the derivative.
sqrt(4-x) in the denominator gives us another zero.

it's a different kind of critical point, but uhh we should probably still include it.

Answer 22

lol :3

Answer 23

Setting the denominator equal to zero,
\[\Large\sf \sqrt{4-x}=0\]What value of x do we get here? :o

Answer 24

4

Answer 25

?

Answer 26

yes :)

Answer 27

would that also be a critical point ?

Answer 28

and also, if you graph this... how would you find that absolute max?

Answer 29

Yes.
It's not the same type of critical point as x=8/3.

It's telling us where the function has a `vertical` tangent which is not as helpful.
But it's still a value we would consider when trying to determine max/min and such.

Answer 30

I guess we would want to determine what type of critical point x=8/3 is. :o

Answer 31

There are several ways to do that.
I like doing this personally.
Drawing out a number line and plugging in a couple of test points to see what's happening on the left and right of our critical number.

Answer 32

right, which ive already solved. so to get that critical point which is also your abs and local max, you ust plug 8/3 to the equation ?

Answer 33

Oh so you can write it as an ordered pair?
Yah that sounds good!
Back into the `original equation`.

Answer 34

ok cool . thanks so much!