Question

Find the derivative algebraically using the difference quotient: q(x) = x^-2 at x=2

Answer 1

is that \[\large q(x)=x^{-2}\]

Answer 2

yes

Answer 3

the diffrence quotient is :
\[\large \frac{f(x)-f(2)}{x-2}\]

Answer 4

that's what you need to find out with limit:
\[\Large \lim_{x\to2}\frac{f(x)-f(2)}{x-2}=?\]

Answer 5

I thought the difference quotient is \[\frac{ f(x+h)-f(x) }{ h } \]

Answer 6

well that's the same thing my friend

Answer 7

\[I got this far but I can't solve the rest: \frac{ -4-h }{ 4h^2+6h+16 }\]

Answer 8

if you want that way it would be \[\large \lim_{h\to0}\frac{f(2+h)-f(2)}{h}\]

Answer 9

what did you use?

Answer 10

But it is x^-2 not x^2

Answer 11

i'm giving the definition didn't do anything yet my friend

Answer 12

oh, okay

Answer 13

you are miss using the definition

Answer 14

try to use the one i just wrote

Answer 15

using what you gave me I got: \[\frac{ \frac{ 1 }{ (2+h)^2 }-\frac{ 1 }{ 4 } }{ h }\]

Answer 16

okay simplify

Answer 17

\[\frac{ \frac{ -(4+4h+h^2)+4 }{ (4+4h+h^2)4 } }{ h }\]

Answer 18

go further...

Answer 19

carry some cancelations

Answer 20

that is why I typed that thing earlier
I simplified and got to that point and got stuck

Answer 21

well still need to go further... open perenthesis on top and see what can yu spot

Answer 22

\[Simplified \to \frac{ -4-h }{ 16+6h+4h^2 }\]

Answer 23

algebra algebra all is algebra...

Answer 24

o don't think so forgot that part and thing about the one you typed just now

Answer 25

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

Answer 26

that's the first thing to spot

Answer 27

see what you can cancel

Answer 28

@misty1212 pls let him do this :)

Answer 29

-4h-h^2

Answer 30

h(4-h)

Answer 31

not quite h(-4-h)

Answer 32

no cancel h top and bottom

Answer 33

I've gotten to that part

Answer 34

you got incorrect math in there

Answer 35

\[\large \frac{-4-h}{4(2+h)^2}\] oh i see you got to this point

Answer 36

well you need to bring that limit

Answer 37

this is jut difference quotient
we need limit to make it derivative

Answer 38

so take the limit:
\[\large \lim_{h\to0}\frac{-4-h}{4(2+h)^2}\]

Answer 39

When I canceled the h from the top and bottom I got:
\[\frac{ -4-h }{ (4+4+h^2)4 }\]

Answer 40

wait
h not h^2

Answer 41

no cancelation anymore
do the limit i just wrote
i said it appears you got the right thing from the beginning

Answer 42

you just had some errors, some number were not correct in denominator

Answer 43

so I don't have to cancel anymore?

Answer 44

no further cancelation there is nothing to cancel anymore
just do the limit

Answer 45

oh okay, I see now...So I did have it right earlier I just needed to find the limit?

Answer 46

yes sort of, you got the number in the bottom incorrect
but other than that it is fine

Answer 47

so the limit is -1/4

Answer 48

\[\lim_{h\to0}\frac{-4-h}{4(2+h)^2}\] you did this limit?

Answer 49

yes

Answer 50

how its -1/4?

Answer 51

oh i'm blind hehe you are correct

Answer 52

forgot to the power 2 lol

Answer 53

\[\frac{ -4-(0) }{ 4(2+0)^2 }\]

Answer 54

yes its correct :) i was not paying attention enough

Answer 55

sigh...so did I do it correctly before when I wrote
\[\frac{ -4-h }{ 16+6h+4h^2 }\]

Answer 56

well the bottom was 4(2+h)^2=16+16h+4h^2

Answer 57

i said you got a math error and that's fine

Answer 58

oh I see where I got the error

Answer 59

I'm lagging sorry the website is bad again hehe

Answer 60

I have it written correctly on paper but when I translated it to the computer I wrote 16 instead of 6

Answer 61

Okay.. thank you

Answer 62

we can check if our result is good:
\[\large f'(x)=-2x^{-3}=-2/x^3\]
so \[\large f'(2)=-2/8=-1/4\]

Answer 63

you are welcome!