Question

Use the definition of the derivative f'(c) to find f'(c) where f(x)=sqrtx +2x and c=1

I just need to see how the function is set up and I can figure it out from there.

Answer 1

\[\large\rm f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}\]They want us to use c=1,\[\large\rm f'(1)=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}\]


\[\large\rm f(\color{orangered}{x})=\sqrt{\color{orangered}{x}}+2\color{orangered}{x}\]At x=1,\[\large\rm f(\color{orangered}{1})=\sqrt{\color{orangered}{1}}+2\color{orangered}{\cdot1}\]and x=1+h,\[\large\rm f(\color{orangered}{1+h})=\sqrt{\color{orangered}{(1+h)}}+2\color{orangered}{(1+h)}\]

These are the two pieces that will get plugged into your numerator.

Answer 2

okay so how would you solve that per say since I plugged it all in and got 2 and Im not sure if thats correct

Answer 3

Well, after you plug everything in,\[\large\rm f'(1)=\lim_{h\to0}\frac{\sqrt{1+h}+2(1+h)-(\sqrt1+2)}{h}\]The numerator should simplify a little bit, ya?\[\large\rm f'(1)=\lim_{h\to0}\frac{\sqrt{1+h}+2h-1}{h}\]

Answer 4

You got 2? 0_o hmm

Answer 5

I didnt get the simplified version so I did something wrong

Answer 6

There is a bunch more simplification,
it's going to feel sort of weird.

Answer 7

I keep cancelling out the h's so it keeps equally 2 so i get sqrt1 +2-1 so Im clearly doing something wrong

Answer 8

\[\large\rm f'(1)=\lim_{h\to0}\frac{\sqrt{1+h}+2h-1}{h}\]Direct substitution does not seem to work from this point, ya?
We still have that pesky h in the bottom.
Let's split the numerator into two groups,\[\large\rm f'(1)=\lim_{h\to0}\frac{\sqrt{1+h}+(2h-1)}{h}\]To get rid of that square root, we can apply conjugation:
Multiplying the numerator and denominator by the conjugate of the numerator.

Answer 9

so 1+h+(2h-1)^2/h^2?

Answer 10

No no.
We're not multiplying like this,\[\large\rm f'(1)\ne\lim_{h\to0}\frac{\sqrt{1+h}+(2h-1)}{h}\color{royalblue}{\left(\frac{\sqrt{1+h}-(2h-1)}{h}\right)}\]we're multiplying like this,\[\large\rm f'(1)=\lim_{h\to0}\frac{\sqrt{1+h}+(2h-1)}{h}\color{royalblue}{\left(\frac{\sqrt{1+h}-(2h-1)}{\sqrt{1+h}-(2h-1)}\right)}\]

Answer 11

So we shouldn't be getting the h^2 in the bottom, ya?

Answer 12

And remember,
multiplying conjugates gives you the `difference` of squares.
So we should be getting subtraction between the split that we made.

Answer 13

oh you mean the entire thing oooooh okay i see where you were going with that. means that what would it look like after you multiply it since Im really lost at this point

Answer 14

Here's a hint when applying this conjugate trick.
`Only expand out the conjugates.`
So in this case, it might be tempting to multiply the h in the denominator by that big mess, but don't! Just leave the multiplication sitting there. Don't expand the bottom.\[\large\rm f'(1)=\lim_{h\to0}\frac{(1+h)-(2h-1)^2}{h[\sqrt{1+h}-(2h-1)]}\]Your numerator was very close to being correct.
You just had the wrong sign between them.

Answer 15

Simplify the numerator, what do you get?

Answer 16

(1+h)-(2h-1)/h[sqrt1+h ] ?

Answer 17

? I don't understand what you did..

Answer 18

Oh oh I think I see what you're trying to do.
No no, don't ever do that again :D lol

Answer 19

\[\large\rm \frac{a+b}{c+b}\ne\frac{a+\cancel b}{c+\cancel b}\]bad bad bad.
You can not do this.

Answer 20

I don't either honestly. You see I don't know what I am doing since my class is extremely vague so yea I apologise for now knowing. Also I kinda get why I cant do that but you see that how they taught us so I do it automatically

Answer 21

Simplify the numerator,\[\large\rm (1+h)-(2h-1)^2=?\]

Answer 22

1+h-4h-4h^2-1?

Answer 23

it should be +1 sorry

Answer 24

I don't think you distributed the negative properly.
Or you didn't altogether? Hmm

Answer 25

\[\large\rm (1+h)-(2h-1)^2\]expanding the square,\[\large\rm (1+h)-(4h^2-4h+1)\]

Answer 26

Oh geez I give up, I have another problem I have to do and that the same as this one and I still dont know what Im doing and oh yea I just combined it thats all

Answer 27

Distributing the negative gives us,\[\large\rm (1+h)-4h^2+4h-1\]

Answer 28

oh okay I see. do we have to factor it?

Answer 29

Combine like-terms,\[\large\rm -4h^2+5h\]That should be all that remains.

Answer 30

In the numerator*

Answer 31

ah I see so would based on all of that, would the answer be 1/2?

Answer 32

It's still too early to be guessing answers.
Not sure why you're doing that :P Hmm

Answer 33

sorry its a multiple choice question so Im jsut guessing

Answer 34

\[\large\rm f'(1)=\lim_{h\to0}\frac{(1+h)-(2h-1)^2}{h[\sqrt{1+h}-(2h-1)]}\]So our numerator expanded and simplified to give us,\[\large\rm f'(1)=\lim_{h\to0}\frac{4h^2-5h}{h[\sqrt{1+h}-(2h-1)]}\]We should probably pull an h out of each term in the numerator,\[\large\rm f'(1)=\lim_{h\to0}\frac{h(4h-5)}{h[\sqrt{1+h}-(2h-1)]}\]Ooo now we can get rid of that pesky h in the denominator!
Do you see it?

Answer 35

Oh sorry it's (4h+5) in the numerator.

Answer 36

okay yes I see that but how would we get rid of the rest? would we have to multiple it to remove the sqrt

Answer 37

Well we have this nice cancellation happening,\[\large\rm f'(1)=\lim_{h\to0}\frac{\cancel h(4h+5)}{\cancel h[\sqrt{1+h}-(2h-1)]}\]In limit problems, any time you're able to mke a cancellation, go back to your first line of attack: Direct substitution.

Something has changed, and we need to check and see if direct substitution will work from this point.

Answer 38

\[\large\rm f'(1)=\lim_{h\to0}\frac{4h+5}{\sqrt{1+h}-(2h-1)}\]So what does direct substitution give you?
Replacing h with 0.

Answer 39

5/1-(-1)=5/2?

Answer 40

5/(1-(-1))=5/2?

Yes, good job.
That's our final answer.

Answer 41

ahhhh i see cool thank you for helping me with that now I have one more thats just like this one, but f(x)=sqrt(x+2) -3 and c=-1. Can you help me with this one as well? Its my last problem.

Answer 42

This one will probably be a tad easier since we don't have multiple x's in the function.

Answer 43

oh yea thank you

Answer 44

\[\large\rm f'(-1)=\lim_{h\to0}\frac{\color{royalblue}{\sqrt{x+h+2}-3}-\color{orangered}{(\sqrt{x+2}-3)}}{h}\]Here is our limit definition, with the two pieces plugged into the numerator.

Answer 45

Oh we're evaluating this at x=-1, woops.

Answer 46

\[\large\rm f'(-1)=\lim_{h\to0}\frac{\color{royalblue}{\sqrt{-1+h+2}-3}-\color{orangered}{(\sqrt{-1+2}-3)}}{h}\]

Answer 47

okay I have that all written down

Answer 48

so would we multiply it by the numerator to get rid of the sqrt?

Answer 49

Let's simplify a little before moving too far.
2 and -1 gives us 1. Square root of 1 is 1.\[\large\rm f'(-1)=\lim_{h\to0}\frac{\color{royalblue}{\sqrt{1+h}-3}-\color{orangered}{(1-3)}}{h}\]We can simplify further still, ya?

Answer 50

sqrt1+5-1?

Answer 51

sorry I didnt meant to put the 5 in there

Answer 52

sqrt1+h -1

Answer 53

sqrt(1+h)-1 ok good.

Answer 54

That's really unclear when you don't write it with brackets :) lol

Answer 55

ah sorry ill try to do that

Answer 56

\[\large\rm f'(-1)=\lim_{h\to0}\frac{\sqrt{1+h}-1}{h}\]Ok conjugation from this step seems like a good idea :)

Answer 57

so 1+h+1/h sqrt(1+h) -1

Answer 58

Multiplying conjugates should give you the `difference` of squares.
So we shouldn't have the second addition in the numerator.
(1+h)-1

And what's going on in the denominator?
Again, you need to learn to use brackets...
h(sqrt(1+h)-1)
But this is still a little off.

We multiplied top and bottom by sqrt(1+h)+1), right?

Answer 59

ah right okay so let me write that out to visually it.

Answer 60

\[\large\rm f'(-1)=\lim_{h\to0}\frac{\sqrt{1+h}-1}{h}\color{royalblue}{\left(\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1}\right)}\]Ya, it uses a lot of paper to write it out, but I like to see that step when I'm working. :)

Answer 61

so (1+h)-1/h(sqrt(1+h)+1)

Answer 62

Ok great.
We can probably drop the brackets on the (1+h) since nothing is being applied to them,\[\large\rm f'(-1)=\lim_{h\to0}\frac{1+h-1}{h(\sqrt{1+h}+1)}\]and then simplify the numerator further.

Answer 63

so just h?

Answer 64

Just h left up top?
Ok good.
Cancel some h's.

Answer 65

\[\large\rm f'(-1)=\lim_{h\to0}\frac{\cancel h}{\cancel h(\sqrt{1+h}+1)}\]Gotta be careful when using this idea of "cancelling",
think for a sec, what is left in the numerator?

Answer 66

1?

Answer 67

Good good good.
It's not zero.

Answer 68

\[\large\rm f'(-1)=\lim_{h\to0}\frac{1}{\sqrt{1+h}+1}\]And once again, we can try direct substitution.

Answer 69

okay let me do that now

Answer 70

1/2?

Answer 71

Yay good job

Answer 72

cat cat kitty cat meow meow,
practice those Algebra skills!

Answer 73

Lol thank you oh so much! I had about three mental breakdowns trying to do this and you just helped me get to the finish line. I really appreciate it.

Answer 74

np
take a break, drink a beer
unless you're not old enough to drink XD then dont

Answer 75

Haha lol will do (except the drinking, I'm not quite there yet) and thank you and I appreciate the gesture.