Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

\[\frac{ \sqrt{2(x+h)+6}-\sqrt{2x+6} }{ h }*\frac{ \sqrt{2(x+h)+6}+\sqrt{2x+6} }{ \sqrt{2(x+h)+6} +\sqrt{2x+6}}\]
\[\frac{ 2(x+h)+6-2x-6 }{ h(\sqrt{2(x+h)+6}+\sqrt{2x+6}) }\]
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.

Answer 6

WAIT OKAY NOW I SERIOUSLY NEED YOUR HELP

Answer 7

What in the hell is f '(x)

Answer 8

http://gyazo.com/6e93cdf22d480a72829e6023213aa3ba

Answer 9

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

Answer 10

Well then. I have never learned derivatives :O

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

Its 5x :P

Answer 19

Well that was easy..

Answer 20

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

Answer 21

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.

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

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?

Answer 29

Looks good.

Answer 30

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

Answer 31

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

Answer 32

set h = to 0 now.

Answer 33

-7+4x?

Answer 34

Yep, thats the derivative.

Answer 35

http://gyazo.com/4971d14bba106e7c898806e9b7f63cdd

Answer 36

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

Answer 37

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

Answer 38

You can show me! :D

Answer 39

\[f(x) = x ^{n}\implies f'(x) = nx ^{n-1} \]

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.

Answer 40

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

Answer 41

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.

Answer 42

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

Answer 43

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

Answer 44

oh sorry I mean m!

Answer 45

Okay so the deriv. is 2x

Answer 46

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

Answer 47

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

Answer 48

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.

Answer 49

Where do you live?

Answer 50

west coast US.

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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.

Answer 55

Hmm..

Answer 56

Got it:)

Answer 57

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

Answer 58

NOOOOOOOOOOO:(

Answer 59

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

Answer 60

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

Answer 61

Thank you so much for your help