Question

Find the instantaneous rate of change of y with respect to x at an arbitrary value of x(v not)

Answer 1

\[y=\frac{ 1 }{ x^2 }\]
\[x_{o}=1\]
\[x_{1}=\]

Answer 2

\[x_{1}=2\]

Answer 3

what is the x(v not)?

Is that supposed to be a caret pointing down? like subscript? :)
Use underscore for that.
So we want to find y' evaluated at x_naut?
Oh and they gave us a value for that? neato

Answer 4

yea.

Answer 5

Let's use rules of exponents to fix our function up a little bit.\[\Large y=\frac{1}{x^2}\qquad\to\qquad y=x^{-2}\]

Answer 6

Have you learned about the `Power Rule` for derivatives?

Answer 7

no, not yet.

Answer 8

So we're using the `Limit Definition of the Derivative` to solve this?

Answer 9

I am just starting derivatives, in the first section of the chapter, the ones with the slope and formula for tangent and secant

Answer 10

yeah if u call it that

Answer 11

So here is our thing we'll use:\[\Large y'=\lim_{h\to0}\frac{y(x+h)-y(x)}{x}\]

I'm writing y in function notation, hope that's not too confusing :o

Answer 12

AHHH typo, lemme fix that

Answer 13

\[\Large y'=\lim_{h\to0}\frac{y(x+h)-y(x)}{h}\]

Answer 14

You're probably more familiar with seeing f(x+h)-f(x) on top.
Is it too confusing doing it this way?

Answer 15

no its ok, but is it possible to use this one,\[y ^{'}=\lim_{x _{1} \rightarrow x _{o}}\frac{ f(x _{1})-f(x _{o}) }{ x _{1}-x _{o} }\]

Answer 16

im not really accustomed to the h, but the teacher has showed us that eqtn

Answer 17

Ah yes that form :) that would make more sense.

Answer 18

mm one sec :3

Answer 19

sure, no problem

Answer 20

I don't think we want to use both initial x values for that form.
The other limit formula, (the one you're used to using), is usually written as:\[\LARGE f'(x_o)=\lim_{x\to x_o}\frac{ f(x)-f(x _{o})}{x-x_o}\]Where we let the first x vary.

Answer 21

Sorry I haven't used that limit formula very often lol :) maybe i'm being silly, lemme work it out on paper real quick.

Answer 22

the first x meaning \[x _{o}\]?

Answer 23

The first x meaning x. As in, leave it as a variable, plug it in and simplify the thing down.
Then after we've gotten it into a nice form, we'll let "x approach x_o". (plug in x_o for our x's).

Answer 24

It's really extra confusing that we're using y AND f(x) notation.

Does this make sense on the left side of the equation if I write: \[\LARGE f'(x_o)=\lim_{x\to x_o}\frac{ f(x)-f(x _{o})}{x-x_o}\]
Instead of y'?
In which case we should write our original function as:\[\Large f(x)=\frac{1}{x^2}\]

Answer 25

\[\Large \color{royalblue}{f(x)=\frac{1}{x^2}}\]\[\Large \color{orangered}{f(x_o)=\frac{1}{x_o^2}\qquad\to\qquad f(1)=\frac{1}{1^2}=1}\]
We want to use these two pieces of information and plug them into our limit.

\[\LARGE f'(x_o)=\lim_{x\to x_o}\frac{ \color{royalblue}{f(x)}-\color{orangered}{f(x _{o})}}{x-x_o}\]

Answer 26

I color-coded them, maybe it will help match things up D:

Answer 27

Understand how we're going to plug those pieces in?
Or still way too confused on the limit formua that I used? :3

Answer 28

it makes sense but the answer i got for this didnt match the answer key, no matter how i manipulated it, i was still alittle off from the book's answer.

Answer 29

Plugging in those two pieces of information gives us this so far:\[\Large \lim_{x\to x_o}\frac{ \color{royalblue}{\dfrac{1}{x^2}}-\color{orangered}{1}}{x-1}\]

Answer 30

Getting a common denominator up top gives us,\[\Large \lim_{x\to x_o}\frac{\left(\dfrac{1-x^2}{x^2}\right)}{x-1}\]

Answer 31

Oh we should have a 1 in place of that x_o under our limit also.

Answer 32

We can combine our denominators like so:\[\Large \lim_{x\to1}\dfrac{1-x^2}{x^2(x-1)}\]

Answer 33

yes i get it, but that was not the way i did it, but ok

Answer 34

This problem is a tad tricky, a lot of little steps to get there D:

Answer 35

The numerator can be factored into the `difference of squares`:\[\Large \lim_{x\to1}\dfrac{(1-x)(1+x)}{x^2(x-1)}\]
We'll factor out a negative 1 from the (1-x) term.\[\Large \lim_{x\to1}\dfrac{-(x-1)(1+x)}{x^2(x-1)}\]
Which gives us a nice cancellation.

Answer 36

k, that i understand also

Answer 37

\[\Large \lim_{x\to1}\dfrac{-\cancel{(x-1)}(1+x)}{x^2\cancel{(x-1)}}\]From this form we can finally plug in x=1 (allowing x to approach 1).\[\Large \lim_{x\to1}\dfrac{-(1+x)}{x^2}\]

Answer 38

What do you get?? :D

Answer 39

ok, thats getting close

Answer 40

im getting all the algebra ur doing

Answer 41

You should be able to get your final answer by plugging x=1 directly in from that point.

Answer 42

actually that was the solution that i got frist time i did it, then i checked with my teacher and he shrugged and said x naut should be included in your final answer. so therefore i did it again and i got stuck in factoring...

Answer 43

That doesn't make any sense, why would x_o be included in your final answer if they provide a value for it.... that just seems silly -_-

Answer 44

i dunno about that, cuz this is just c) of a problem, maybe reread the question i have up there? i have no idea

Answer 45

If you're supposed to leave x_o in the equation, it would work out largely the same way.\[\Large f'(x_o)\quad=\quad\lim_{x\to x_o}\frac{\dfrac{1}{x^2}-\dfrac{1}{x_o^2}}{x-x_o}\]

Answer 46

Common denominator up top gives us,\[\Large f'(x_o)\quad=\quad\lim_{x\to x_o}\frac{\left(\dfrac{x_o^2-x^2}{x^2x_o^2}\right)}{x-x_o}\]

Answer 47

That step is a little tricky :o understand how to combine those fractions?

Answer 48

omgAAAAD, i was typing so much and the page broke....