Question

(Calculus) I am doing derivatives, and I am completely lost.
First of all, I get the basic part (I think), but...

Answer 1

Why does (according to my book) 3x^(2) + 3xh + h^(2) = 3x^(2) ? I know how to do this the short way, but I am going back over my notes and cant it out

Answer 2

hmmmmm

Answer 3

h = 0

Answer 4

that is what I can come up with
but I think you wrote something inaccurately

Answer 5

derivative is concern about limits
and usually in this case the limit approaches zero but not equal to zero

Answer 6

it came out of my book, I am not quite sure. It is the derivative of x^3

Answer 7

write down the derivative by limit definition

Answer 8

You don't have to do this, but I'd suggest you write out what it is that confuses you here and what you need to know.

For example, it would help you, and help others who may want to help you, if you'd start off by saying, "I'm learning to find the derivative of a function using the DEFINITION of derivative.

So: would you mind stating in words just what you want to know in this particular instance?

Answer 9

the derivative of x^3 is 3 -:(

Answer 10

Oh, I am doing just the definition right now

Answer 11

i'm out, not interrupting anymore

Answer 12

derivative of x^3 follows the power rule
\[\frac{ dy }{ dx } ax^n = anx^{n-1}\]

Answer 13

so x^3 will be 3x^(3-1) right?

Answer 14

Yes, I'd guessed that, but I'd like for you to tell me in words just what you want to know.
@nincompoop : That formula comes later. Here we're dealing with the DEFINITION OF THE DERIVATIVE.

Answer 15

sorry, I though 3x, and it's x^3

Answer 16

ght

Answer 17

Yes, I can get the answer through the power function, but I am having trouble with the definition. f'(x)=[(x+h)^(3) - x^(3)]/h . I followed along with the example, and got to the last part which was 3x^(2) + 3xh + h^(2)

Answer 18

\[\lim_{ h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }\]

Answer 19

meaning you are adding a little bit of x, which in this case represented by h instead of delta x, and we are concern about that a little bit of x that we added.

Answer 20

OK. Thanks for writing this out. The last part involves something very important: taking the limit of your very last result as the parameter h goes to zero.

So, please write all of this (no omissions):\[\lim_{ h \rightarrow 0}\frac{(x+h)^3-x^3 }{ h }\]

Answer 21

Yes, you do need to expand (x+h)^3.

Answer 22

\[\lim_{ h \rightarrow 0}\frac{ x^3+3x^2h +3xh^2+h^3-x^3 }{ h }\]

Answer 23

Notice how the x^3 terms cancel?
What's left in the numerator is an expression that includes both x and h.
Divide every term of that expression by h.
Finally, let h approach zero.

what's left is the derivative you wanted, f '(x) = 3x^2.

I need to get off the 'Net now, but feel confident that you can finish this problem yourself. Good working with you.

Answer 24

you have to factor out h since you are trying to avoid to divide by h, which is zero

Answer 25

i end up with 3x^(2) +3xh +h^(2)

Answer 26

or you can try with a different approach that leads to different understanding of the same concept

Answer 27

whatever you have left that has h will be zero so you are left with 3x^2

Answer 28

Good. But you have not yet let h approach zero, have you?
The first term will be 3x^2, as expected.
The second and third terms both have " h " in them, and so, as h approaches zero, so do those two terms.

So you're left with 3x^2. In other words, the desired derivative is f '(x) = 3x^2.

I assume you're learning the power rule for differentiation now, or will soon be learning it; that is the approach suggested earlier by Nincompoop.

Answer 29

that becomes your general formula for the derivative

Answer 30

Good night, you two. Bedtime.

Answer 31

Thanks mathmale. nincompoop, could you explain mathmale means by letting h approach zero? I think i missed a lecture or something

Answer 32

the limit is set as h -> 0
therefore the value of h has to approach zero

Answer 33

http://tutorial.math.lamar.edu/Classes/CalcI/Tangents_Rates.aspx

Answer 34

http://tutorial.math.lamar.edu/Classes/CalcI/Tangents_Rates_files/image003.gif

Answer 35

dx means a little bit of x
so the relative smallness here is set approach zero, but not exactly zero

Answer 36

I'm headed over there. Thanks for the link. before I start reading, can you tell me if f'(x) always equals x, g'(x) always equals x^(2) ?

Answer 37

no

Answer 38

Ok, good. I am getting myself confused.

Answer 39

it's normal, because the idea is not at first very intuitive

Answer 40

the limit problem is one of the most difficult concept to grasp when starting calculus

Answer 41

Thank you so much for your help. I am starting to get this, but I am in dire need of sleep.

Answer 42

Holy crap! I think i just had my first breakthrough in calculus! My school requires business majors to take Calculus with business applications, so my teacher is leaving a lot out. That link is the first time I saw a tangent line.