Find the simplified difference quotient of f(x)=sqrt(2x+6)

- Lukecrayonz

Find the simplified difference quotient of f(x)=sqrt(2x+6)

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- katieb

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- Lukecrayonz

So basically I just need help with sqrt(2(x+h)+6)

- Psymon

\[\frac{ \sqrt{2(x+h)+6}-\sqrt{2x+6} }{ h } \]
Multiply top and bottom by the conjugate.

- Lukecrayonz

[sqrt(2(x+h)+6)-sqrt(2x+6)/h]*h

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## More answers

- Lukecrayonz

End answer:
2/sqrt(2x+2h+6)+sqrt(2x+6)

- Psymon

\[\frac{ \sqrt{2(x+h)+6}-\sqrt{2x+6} }{ h }*\frac{ \sqrt{2(x+h)+6}+\sqrt{2x+6} }{ \sqrt{2(x+h)+6} +\sqrt{2x+6}}\]
\[\frac{ 2(x+h)+6-2x-6 }{ h(\sqrt{2(x+h)+6}+\sqrt{2x+6}) }\]
You get 2x + 2h + 6 - 2x -6, so everything but 2h cancels, but then the h on bottom cancels out with the one on top, so yeah, looks like you got it, lol. Guess I was workign through it but you got it.

- Lukecrayonz

WAIT OKAY NOW I SERIOUSLY NEED YOUR HELP

- Lukecrayonz

What in the hell is f '(x)

- Lukecrayonz

http://gyazo.com/6e93cdf22d480a72829e6023213aa3ba

- Psymon

f'(x) means derivative of f(x)

- Lukecrayonz

Well then. I have never learned derivatives :O

- Psymon

You have, kinda, but you dont know it. The difference quotient is the "definition" of a derivative. Everytime youve done the difference quotient youve partway found the derivative. To find the derivative, do the differentience quotient and at the very end, make any h's remaining into 0s. What is left after that is the derivative.

- Lukecrayonz

So do I always turn H into zero, or if it f'(x) lim-->1 I input it as 1?

- Lukecrayonz

Sorry if thats confusing, its because the first answer is f'(0)

- Psymon

Well, for a derivative its always as h approaches 0. Limit going to a different number is somethign completely different. Also, derivative must be difference quotient in this case. Youll learn how to do derivatives without difference quotient, but this is just the definition. Differene quotient as h goes to 0 = derivative. Anything else is just something else.
f'(0) means find the derivative and then set x = 0 after that.

- Lukecrayonz

Ahhh okay, I thought we set h to zero because of f ' (0)

- Lukecrayonz

So now the question asks "f'(x)"

- Lukecrayonz

Is that my final answer of the difference quotient? 5x+2.5h or would it be something else

- Lukecrayonz

Its 5x :P

- Lukecrayonz

Well that was easy..

- Lukecrayonz

The slope of a line tangent to f(x)=x^2 at x=5 should be

- Psymon

Well, thats what you use the derivative for. The derivative gives youa formula for slope. So if you took the derivative of x^2, youd get 2x. Then plug in x = 5, and you see that you get 2(5) = 10, meaning the slope of the tangent line at x = 5 is 10.

- Psymon

Well, derivatives can do a bunch of things, but with graphs it helps give ya slope.

- Lukecrayonz

Okay wait, what about for f(x)=5x+3

- Lukecrayonz

5(x+h)+3-(5x+3)
5x+5h+3-5x-3
5h
set h to zero
5?

- Lukecrayonz

So there really isn't any guarantee that theres going to be a variable in the end?

- Psymon

Well, like I just said, the derivative helps you find the slope of a graph. When you want the slope at a certain point, you plug in the x value after you take the derivative. But think of it, this graph is already a line. y = 5x + 3 is slope of 5, y-intercept of 3. So when you take the derivative of course youre going to get 5, the slope on the entire graph is 5.

- Lukecrayonz

f(x)=2x^2-7x+5

- Lukecrayonz

2(x+h)(x+h)-7(x+h)+5
2 h^2+4 h x+2 x^2-7x-7h+5-(2x^2-7x+5)
-7+2 h+4 x?

- Psymon

Looks good.

- Lukecrayonz

I added the divide by h in the second step, didn't show work

- Lukecrayonz

So for my f ' (x), what is my answer?

- Psymon

set h = to 0 now.

- Lukecrayonz

-7+4x?

- Psymon

Yep, thats the derivative.

- Lukecrayonz

http://gyazo.com/4971d14bba106e7c898806e9b7f63cdd

- Lukecrayonz

I'm looking at an example and this just looks so messy.

- Psymon

So an equation for a tangent line is going to be linear. Its going to end up in the old point slope form of y-y1 = m(x-x1)
Now we're given a point, so all we need is the slope. Since the derivative is an equation for slope, what we do is take the derivative then plug in the x-coordinate. That will give us the slope we want. Then that slop plus the point go into the point-slope form equation. They need to show you the fast derivative alreayd for these, they take too long otherwise xD

- Lukecrayonz

You can show me! :D

- Psymon

\[f(x) = x ^{n}\implies f'(x) = nx ^{n-1} \]
Basically, the power of comes down as a multiplication and then the power is reduced by 1. So for example, 4x^3. The power of 3 comes down as a multiplication (3)4x^3. And then the power is reduced by 1, so that leaves you with 12x^2. So this problem is x^2 + 5. So the power of x is 2 and comes down to give us 2x^2. Then the power is reduced by 1 giving us 2x. The 5 has x to the power of 0. So when you bring down a 0 power, the 5 goes away and you only have 2x left (constants by themselves are klled in a derivative). So that means the derivative of x^2+5 is just 2x.

- Lukecrayonz

So wait, take the equation x^2+5, input into differential, and instead of x^2 use 3 for x
so I end up with (3+h)^2+5-(3^2+5)/h

- Psymon

No no no, do the derivative first. Once youre done with the derivative and setting h = 0, you get 2x. Plug 3 into THAT one.

- Lukecrayonz

o.O My book says to do it the way I did. But I'll try your way ^_^ But I got h=-6

- Psymon

h = -6? How did we get h = anything but 0, lol. Its supposed to be limit as h goes to 0.

- Lukecrayonz

oh sorry I mean m!

- Lukecrayonz

Okay so the deriv. is 2x

- Psymon

Right. Then plugging in x = 3, since its the x-coordinate of our point, we get m = 6. So now, given an m of 6 and the point (3,14), we can use the point-slope form formula to get
y-14 = 6(x-3)
y = 6x - 4

- Lukecrayonz

That method you just taught me is literally the coolest thing i've ever learned in math

- Psymon

Yeah, it lets you do a lot of them really fast. YOull learn it soon. The derivatives do get tricky as the functions get tricky, but these ones we have are all very straight forward.

- Lukecrayonz

Where do you live?

- Psymon

west coast US.

- Lukecrayonz

Why are you awake O.O I'm East Coast but still

- Psymon

Dunno. My sleep is always crazy xD Im not up for long, though.

- Lukecrayonz

Hmm okay, I get the way my book says to do it but I'd like to see your method
http://gyazo.com/74c0a9f9eb4d47d9b56aa142fe3f542d

- Psymon

Alrighty. SO -3 - 5x^2. Given the regular formula for derivatives
\[x ^{n}\implies nx^{n-1}\]
Constants will always become 0 when you take the derivative. The reason is a constant is x^0. And when you bring down the 0 to mulltiply, the whole term is 0. So looking at the second term, I bring the power down as a multiplication to get (2)-5x^2. Then I lower the power by 1 to get me -10x. Because the point is (-4,-83), I plug in this x-coordinate into the derivative meaning my slope at the given point is 40. So using the slope of 40 and the given point, I can put
y+83 = 40(x+4)
y = 40x+77
So that would be thr tangent line you need.

- Lukecrayonz

Hmm..

- Lukecrayonz

Got it:)

- Psymon

Alrighty, awesome. Im off to sleep now, lol. Night

- Lukecrayonz

NOOOOOOOOOOO:(

- Lukecrayonz

Gahhh I'll have to find someone else as smart as you

- Psymon

im dead tired, cant stay awake x_x Youll find someone to help. Sorry. Night

- Lukecrayonz

Thank you so much for your help

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