Question

Find the derivative of f(x) = -12x2 + 9x at x = 6.

Answer 1

@satellite73

Answer 2

@MrNood

Answer 3

@ganeshie8

Answer 4

\[12x ^{2}+9x\]

Is that the correct formula?

Answer 5

yes

Answer 6

Sorry - I missed out the - sign at the beginning
\[-12x ^{2}+9x\]

Do you know how to get the derivative of

\[y=-12x ^{2}\]

and

\[y=9x\]

Answer 7

Well usually I would put it into the difference quotient formula, meaning i put (x+h) for every x in the original formula, then simplify, and then add the original formula to it , simplify that further and divide the entire thing by h.

Answer 8

But when I originally did that, I got this

(-33hx-33h-9x)/h

which, I don't think is right

Answer 9

So no I don't think I know how to do that. :C

Answer 10

I am not familiar with the method you are describing

do you know for instance that the derivitive of y=2x is 2

or that the derivative of \[y=x ^{2}\]
is 2x

Answer 11

No, my lesson didn't teach me that.. :(

Answer 12

whats the difference quotient formula you have been using ?

Answer 13

These are my answer choices,

-112.5

-135

-90

-108

Answer 14

I'm sorry - the answer is not difficult - but I do not know the method that oyu are being taught.

This a simple derivative and does not need any special formulae

Answer 15

oh, the lesson only spoke about limits really

Answer 16

could you teach me?

Answer 17

And it is

(limh->0) [f(x+h)-f(x)]/h

Answer 18

@ganeshie8

Answer 19

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

Answer 20

yes but i am not famliiar with the ' symbol,

Answer 21

f(x) = -12x2 + 9x

f(x+h) = ?

Answer 22

-12(x+h)^2+9(x+h)

Answer 23

expand

Answer 24

-12(x+h)(x+h)+9x+9h

Answer 25

-12(x^2+2hx+2h)+9x-h

Answer 26

you should get :
f(x+h) = -12(x^2+2hx+h^2)+9x+9h

Answer 27

-24x^2-24hx-24h^2+9x+9h

Answer 28

f(x+h) = -12(x^2+2hx+h^2)+9x+9h
= -12x^2 - 24hx -12h^2 + 9x + 9h

Answer 29

^^

Answer 30

next work the difference : f(x+h) - f(x)

Answer 31

f(x+h) - f(x) = -12x^2 - 24hx -12h^2 + 9x + 9h - (-12x^2 + 9x)
= ?

Answer 32

-24hx-12h^2+9h

Answer 33

excellent !
plug that value in your earlier difference quotient formula

Answer 34

just over h or adding the original formula to it as well, because I thought we did that

Answer 35

would it be,

[-24hx-12h^2+9h]/h

?

Answer 36

thats the difference quotient,
taking the limit gives u the derivative

Answer 37

\[\large f'(x) = \lim \limits_{h\to 0}\dfrac{-24hx -12h^2+9h}{h}\]

Answer 38

Notice that you can factor out `h` on numerator, and then cancel it with the denominator h

Answer 39

Do it.

Answer 40

Oh sorry alright

Answer 41

wait what hold on sorry

Answer 42

\[\large f'(x) = \lim \limits_{h\to 0}\dfrac{-24hx -12h^2+9h}{h}\]


\[\large f'(x) = \lim \limits_{h\to 0}\dfrac{h(-24x -12h+9)}{h}\]

\[\large f'(x) = \lim \limits_{h\to 0}-24x -12h+9\]

Answer 43

Now, you can take the limit because there is no h in the denominator

Answer 44

OH

Answer 45

simply plugin h = 0 in the final expression

Answer 46

\[\large f'(x) = \lim \limits_{h\to 0}\dfrac{-24hx -12h^2+9h}{h}\]


\[\large f'(x) = \lim \limits_{h\to 0}\dfrac{h(-24x -12h+9)}{h}\]

\[\large f'(x) = \lim \limits_{h\to 0}-24x -12h+9\]

\[\large f'(x) = -24x -12(0)+9\]

Answer 47

WOW
okay
So -153?

Answer 48

\[\large f'(x) = -24x + 9\]

Answer 49

I mean -135

Answer 50

Yep !

Answer 51

good job !!

Answer 52

wow thank you so much I'm glad you helped me C:

Answer 53

np, you're welcome :)