Question

Find the derivative of y=(3x^2)/(x^2+2x-1)^1/2

Answer 1

use quotient rule

Answer 2

Are you on a real computer today?
Can you see the math formatted correctly?
\[\large\rm y=\frac{3x^2}{\sqrt{x^2+2x-1}}\]

Answer 3

I can see it

Answer 4

Oo yay! \c:/

Answer 5

I know right! :)

Answer 6

I usually like to "set up" the quotient rule before actually taking the derivatives.
But whatever works best for you.\[\rm \left(\frac{3x^2}{\sqrt{x^2+2x-1}}\right)'=\frac{\color{royalblue}{\left[3x^2\right]'}\sqrt{x^2+2x-1}-3x^2\color{royalblue}{\left[\sqrt{x^2+2x-1}\right]'}}{\left(\sqrt{x^2+2x-1}\right)^2}\]

Answer 7

Omg...of course its not showing up

Answer 8

you -_-
you and your ways...

Answer 9

I see the first half

Answer 10

get on a computer.
get on that chrome :O

Answer 11

Okay just wait I'm at school!

Answer 12

okay I can see it all

Answer 13

so the derivative of the first part would be 6x times the square root I know that

Answer 14

\[\rm =\frac{\color{orangered}{\left[6x\right]}\sqrt{x^2+2x-1}-3x^2\color{royalblue}{\left[\sqrt{x^2+2x-1}\right]'}}{\left(\sqrt{x^2+2x-1}\right)^2}\]Ok good.

Answer 15

And than idk...all I know is for the next part the x^2+2x-1 stay there because it is the chain rule because its more than just x

Answer 16

Gosh i've never been so confused with math!

Answer 17

Derivative of `square root` is really really useful.
You should try to put this one away in your brain somewhere,
instead of converting to 1/2 power.

\[\large\rm \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}\]
Derivative of square root is simply ... one over two square roots.

Answer 18

But yes, we'll have some chain rule business going on,
giving us some extra stuff in addition to that.

Answer 19

okay the derivative of the square root makes sense

Answer 20

So we can generalize our square root derivative,\[\large\rm \frac{d}{dx}\sqrt{stuff}=\frac{1}{2\sqrt{stuff}}stuff'\]When our inner function is more than just x,
we apply the chain rule like this.

Answer 21

So for our problem,\[\large\rm \frac{d}{dx}\sqrt{x^2+2x-1}=\frac{1}{2\sqrt{x^2+2x-1}}(x^2+2x-1)'\]

Answer 22

And what's the derivative of that last piece? :)

Answer 23

2x+2

Answer 24

Ok great. Let's put it in the numerator,\[\large\rm \frac{d}{dx}\sqrt{x^2+2x-1}=\frac{(2x+2)}{2\sqrt{x^2+2x-1}}\]And then let's divide a 2 out of everything.\[\large\rm \frac{d}{dx}\sqrt{x^2+2x-1}=\frac{x+1}{\sqrt{x^2+2x-1}}\]

Answer 25

im confused lol

Answer 26

With what part? :)
Chain rule?
Fancy fraction trick?
The 2 division?
All of it? :d

Answer 27

oh just kidding you took out the 2 on the top and than you cancelled the 2 and you were left with (x+1) on the top?

Answer 28

Ya sorry, I should have included the intermediate step of factoring.
2x+2 = 2(x+1)
and then cancelled the 2's.

Answer 29

nope thats okay, I am a blonde so it can take me a while

Answer 30

\[\rm =\frac{\color{orangered}{\left[6x\right]}\sqrt{x^2+2x-1}-3x^2\color{royalblue}{\left[\sqrt{x^2+2x-1}\right]'}}{\left(\sqrt{x^2+2x-1}\right)^2}\]Mmm k good, that takes care of our other blue piece for us,\[\rm =\frac{\color{orangered}{\left[6x\right]}\sqrt{x^2+2x-1}-3x^2\color{orangered}{\left[\dfrac{x+1}{\sqrt{x^2+2x-1}}\right]}}{\left(\sqrt{x^2+2x-1}\right)^2}\]

Answer 31

Looks like we can do something with the denominator, ya?
Do you see how that will simplify?

Answer 32

it will be just what it is without the square root

Answer 33

wait hold on a minute I got rid of the -3x^2 for some reason?

Answer 34

Yes, square root and square undo one another :)\[\rm =\frac{6x\sqrt{x^2+2x-1}\quad-\quad\dfrac{3x^2(x+1)}{\sqrt{x^2+2x-1}}}{x^2+2x-1}\]

Answer 35

you did? Hmm that's no bueno :O
Try to backtrack and see what happened to it.

Answer 36

I still had it in there I just lost it for some odd reason, and lol love the espanol

Answer 37

lol :D

Ok so now... let's clean it up a tad bit.
We'll get rid of the fraction on fraction business by multiplying through by the "upper" denominator.\[\rm =\frac{6x\sqrt{x^2+2x-1}\quad-\quad\dfrac{3x^2(x+1)}{\sqrt{x^2+2x-1}}}{x^2+2x-1}\left(\frac{\sqrt{x^2+2x-1}}{\sqrt{x^2+2x-1}}\right)\]

Answer 38

And than the square roots can go away?

Answer 39

wait what?

Answer 40

The one in the denominator? Yes.
That's the point of applying this trick.

Lemme add some color, just to make sure it's super clear.

Answer 41

\[\rm =\frac{6x\sqrt{x^2+2x-1}\quad-\quad\dfrac{3x^2(x+1)}{\sqrt{x^2+2x-1}}}{x^2+2x-1}\left(\frac{\color{orangered}{\sqrt{x^2+2x-1}}}{\sqrt{x^2+2x-1}}\right)\]This orange one will be distributed to each term in the numerator.
The other one will simply go to the bottom.\[\rm =\frac{6x\sqrt{x^2+2x-1}\color{orangered}{\sqrt{x^2+2x-1}}\quad-\quad\color{orangered}{\sqrt{x^2+2x-1}}\dfrac{3x^2(x+1)}{\sqrt{x^2+2x-1}}}{(x^2+2x-1)\sqrt{x^2+2x-1}}\]

Answer 42

Uh oh, I smell brain.
Brain esplode?
I hate when that happens :(

Answer 43

I'm so confused

Answer 44

so...? Do the square roots cancel in the second term?

Answer 45

Yes, they cancel :)
That's why we're doing this trick.
To get rid of that nasty fraction.

Answer 46

okay that makes sence now

Answer 47

\[\large\rm =\frac{6x\sqrt{x^2+2x-1}\color{orangered}{\sqrt{x^2+2x-1}}\quad-\quad3x^2(x+1)}{(x^2+2x-1)\sqrt{x^2+2x-1}}\]So yah that whole mess cleans up really really nicely in the second term.

Answer 48

and than the square roots cancel again? hmmm? Am I understanding this right?

Answer 49

ohhhhhhh the square root is multiplied by both terms?

Answer 50

In the numerator?
Each `term` got one of the orange things.
The one with the 3x^2(x+1) cancelled out.

So now only the first term has one.

Answer 51

And ya, the square roots will go away, sort of :)
We'll have to do something clever in the denominator though.

Answer 52

\[\large\rm \sqrt{stuff}\cdot\sqrt{stuff}=stuff\]It just sort of goes away in the numerator though, ya?

Answer 53

Ohhhhhhhhh, this is to much work

Answer 54

so your stuck with 6x(x^2+2x-1)-3x^2(x+1) on the top?

Answer 55

\[\large\rm =\frac{6x(x^2+2x-1)\quad-\quad3x^2(x+1)}{(x^2+2x-1)\sqrt{x^2+2x-1}}\]Mmm looks good.

How bout in the bottom?
It actually benefits us to write this root as a 1/2 power at this point,\[\large\rm =\frac{6x(x^2+2x-1)\quad-\quad3x^2(x+1)}{(x^2+2x-1)^1(x^2+2x-1)^{1/2}}\]

Answer 56

Do you remember how to simplify something like this?\[\large\rm x^1\cdot x^{1/2}\]It requires your exponent rules.

Answer 57

so distribute on the top? and than for the bottom don't you have to add the exponents?

Answer 58

I'm gonna have to go for a bit I'll be back in about 10-15 mins

Answer 59

Good good good :)
You add the exponents.\[\large\rm =\frac{6x(x^2+2x-1)\quad-\quad3x^2(x+1)}{(x^2+2x-1)^{3/2}}\]

Answer 60

And then yes, distribute on the top,
combine like-terms.

Your teacher is really mean by the way,
giving you problems like this.

It's just... wrong.
not cool prof.

Answer 61

I know right, uff I glad I made it home safe, and it is really hard