Question

I really don't know what I am doing so if anyone can answer this question and show their work I would greatly appreciate it.
Find the derivative of f(x) = 8 divided by x at x = -1

Answer 1

you knw the derivative of \(x^n\) ?

Answer 2

no

Answer 3

okay, you want to find it using limits then ?

Answer 4

im pretty sure that that is what i am supposed to do in this lesson

Answer 5

good :)

Answer 6

i am taking an online course and reading the textbook isnt helping me at all

Answer 7

\(\large f'(a) = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x - a}\)


seen this derivative definition before right ?

Answer 8

i haven't seen that before

Answer 9

\(\large f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

wat about this one ?

Answer 10

that one I have seen before, I answered a question using it earlier but for some reason I just haven't been able to apply it to the equation

Answer 11

lets see

Answer 12

but even then, that is still not in the lesson that corresponds to the activity that i am doing

Answer 13

\(\large f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\frac{8}{x+h} - \frac{8}{x}}{h}\)

Answer 14

ive just plugged in f(x+h) and f(x) values above, fine so far ?

Answer 15

yeah

Answer 16

lets get rid of fractions in the numerator, by multiplying top and bottom wid x(x+h)

Answer 17

\(\large f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\frac{8}{x+h} - \frac{8}{x}}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\color{red}{x(x+h)} (\frac{8}{x+h} - \frac{8}{x})}{\color{red}{x(x+h)}h}\)

Answer 18

still fine ? :)

Answer 19

yeah

Answer 20

thanks for checking ^_^

Answer 21

\(\large f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\frac{8}{x+h} - \frac{8}{x}}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\color{red}{x(x+h)} (\frac{8}{x+h} - \frac{8}{x})}{\color{red}{x(x+h)}h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{8\color{red}{x} - 8\color{red}{(x+h)}}{\color{red}{x(x+h)}h}\)

Answer 22

\(\large f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\frac{8}{x+h} - \frac{8}{x}}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\color{red}{x(x+h)} (\frac{8}{x+h} - \frac{8}{x})}{\color{red}{x(x+h)}h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{8\color{red}{x} - 8\color{red}{(x+h)}}{\color{red}{x(x+h)}h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{8\color{red}{x} \color{red}{-8x- 8h}}{\color{red}{x(x+h)}h}\)

Answer 23

\(\large f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\frac{8}{x+h} - \frac{8}{x}}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\color{red}{x(x+h)} (\frac{8}{x+h} - \frac{8}{x})}{\color{red}{x(x+h)}h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{8\color{red}{x} - 8\color{red}{(x+h)}}{\color{red}{x(x+h)}h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{ \color{red}{- 8h}}{\color{red}{x(x+h)}h}\)

Answer 24

\(\large f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\frac{8}{x+h} - \frac{8}{x}}{h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{\color{red}{x(x+h)} (\frac{8}{x+h} - \frac{8}{x})}{\color{red}{x(x+h)}h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{8\color{red}{x} - 8\color{red}{(x+h)}}{\color{red}{x(x+h)}h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{ \color{red}{- 8h}}{\color{red}{x(x+h)}h}\)

\(\large f'(x) = \lim \limits_{h \to 0} \frac{ \color{red}{- 8}}{\color{red}{x(x+h)}}\)

\(\large f'(x) = \frac{ \color{red}{- 8}}{\color{red}{x(x+0)}}\)

\(\large f'(x) = \frac{ \color{red}{- 8}}{\color{red}{x^2}}\)

Answer 25

we're almost done.
let me knw once u digest above

Answer 26

wait so h=0?

Answer 27

the person who helped me before had me solve h first and then plug it into the equation

Answer 28

nope. you dont solve for h as such,
i had a look at ur previous question. he did it exactly the same way we're doing :o

Answer 29

well then im thoroughly confused then lol

Answer 30

in the end you need to plugin h = 0 though

Answer 31

its okay lol, these things are confusing when u just start.... u need to do them over and over again... until u get some notion of how these are done :)

Answer 32

okay, so we got

\(\large f'(x) = \frac{ \color{red}{- 8}}{\color{red}{x^2}}\)

just plug x = -1 to find \(f'(-1)\)

Answer 33

\(\large f'(x) = \frac{ \color{red}{- 8}}{\color{red}{x^2}}\)

\(\large f'(-1) = \frac{ \color{red}{- 8}}{\color{red}{(-1)^2}}\)

\(\large = \color{Red}{- 8} \)

Answer 34

okay. i am going to try a similar problem, would you mind sticking around and making sure i do it right?

Answer 35

sure :)

Answer 36

im going to work step by step like you did so its easy to fix any errors

Answer 37

okay :)

btw if u watch this video, u wil change ur opinion about monkeys i believe :o
http://www.youtube.com/watch?v=5qqdovHOgvU

Answer 38

so the new question is
Find the derivative of f(x) = -11/x at x = 9.

Answer 39

ill watch the video after i conquer this

Answer 40

yup watch it when you're free

Answer 41

(f(x+h)-f(x))/h
(f(-11/(x+h)-f(-11/x))/h

Answer 42

holdup

Answer 43

k

Answer 44

(f(x+h)-f(x))/h

( -11/(x+h)- (-11/x) )/h

Answer 45

now that correct above,
you need not show f() in second step ok

Answer 46

ok

Answer 47

keep going

Answer 48

so then you need to get rid of the denominator so you multiply by it
x(x+h)*( -11/(x+h)- (-11/x) )/h*x(x+h)
-11x+11(x+h)

Answer 49

*(11x+11(x+h))/hx(x+h)

Answer 50

forgot to write the denominator

Answer 51

looks good, continue

Answer 52

wait a sec

Answer 53

*(\(\color{red}{-}\)11x+11(x+h))/hx(x+h)

Answer 54

you missed that -

Answer 55

my bad

Answer 56

simplify numerator, and cancel things u can :)

Answer 57

so then it becomes 11h/hx(x+h)
11/x(x+h)

Answer 58

yup !
now u can take the limit.
plugin h = 0

Answer 59

11/(9)^2
11/81

Answer 60

yay

Answer 61

\(\large \color{red}{\checkmark}\)

Answer 62

good work !!

Answer 63

is it okay if i do 1 more but instead of doing 1 step at a time u just check it after?

Answer 64

sure, go ahead

Answer 65

<3

Answer 66

this is what i have so far

Answer 67

you flipped the sign, \(\color{red}{-}\)12(x+h)^2
in ur second step

Answer 68

crap

Answer 69

give me a sec
ill fix it

Answer 70

:) once u correct, u can get rid of h in the denominator

Answer 71

okie

Answer 72

it leaves me with an h still
i dont have to get rid of it completely right?

Answer 73

yes, we just need to get rid of it in the bottom
as setting h = 0, makes it 0 in the bottom... which is illegal in math

Answer 74

so then i make h=0 after and make x=6 right?

Answer 75

u flipped sign again for 9h

Answer 76

fak

Answer 77

the mistake is on second line from bottom
left side..

Answer 78

i see it

Answer 79

good :)

Answer 80

in the end, you should get

-3(8x+4h-3)

Answer 81

after that plugin, h = 0

Answer 82

in the end i got -120

Answer 83

but i know im wrong because it is not one of the answer choices

Answer 84

whats -3(45) ?

Answer 85

derp

Answer 86

-135

Answer 87

which is an answer choice

Answer 88

I cant say anything but thank you

Answer 89

i would've been completely lost without your help

Answer 90

i need to go to bed so that i stop making a bunch of silly mistakes :)

Answer 91

im sure you deserve some good sleep after so much hard work. sleep tight ! have horrible calculus dreams lol :)

Answer 92

lol thanks again

Answer 93

and have a good night.