Question

Help with Derivatives? :3
Find the derivative of f(x) = -2x^2 + 11x at x = 9.

Answer 1

I have four questions....

Answer 2

@sourwing

Answer 3

use the power rule

Answer 4

d/dx x^n = n x^(n-1)

Answer 5

?

Answer 6

d/dx -2x^2 = -2 (2 x) = -4x
d/dx 11x = 11

so f'(x)= -4x + 11

Answer 7

f(x) = -2x^2 + 11x
If you want to take the derivative of this:
d/dx (f(x) = d/dx ( -2x^2 + 11x)
It is equivalent of taking the derivative of each individual term. That is where we need power rule.

Answer 8

plug in x = 9 and evaluate

Answer 9

What is d?

Answer 10

@phi Can you help me out with this? I know they were trying to help, but I had no idea what they were talking about and then they left. v_v

Answer 11

How much do you know about taking the derivative of
\[ \frac{d}{dx}( -2x^2) \]
?

Answer 12

I didn't learn that way. My school taught us to do the f(x+h)-f(x)/h way.. That doesn't work for this one I don't think...

Answer 13

If you need more info on what that means and how to do it, first try this
http://www.khanacademy.org/math/differential-calculus/taking-derivatives/derivative_intro/v/calculus--derivatives-2-5--new-hd-version

Answer 14

I did it that way, and I didn't have an x left, only an h, and all of my answer choices are just numbers..

Answer 15

Ugh, that time I got 36. Not right either...

Answer 16

ok, you are still learning the basics. once you know them, they teach you the short-cuts.

Answer 17

I'm at ((-2h^2)-4hx)/(h)

Answer 18

\[ f(x)= -2x^2 + 11x \\

f(x+h)= -2(x+h)^2 +11(x+h) \]
we use those in the definition
\[ \lim_{h \rightarrow 0} \frac{ f(x+h)- f(x)}{h}\]

Answer 19

Ohh, I have to do the 11's x too? That was probably where I screwed up..

Answer 20

Here is what I get
\[ f(x+h)= -2(x+h)^2 +11(x+h) \\
= -2(x^2 +2xh +h^2) +11x +11h \\
= -2x^2 -4xh -2h^2 +11x + 11 h
\]
when we subtract f(x) we get
\[ -4xh -2h^2 + 11 h \]

the next step is divide by h

Answer 21

So we get -4x-2h+11.

Answer 22

Doesn't the h disappear?

Answer 23

yes, you get
\[ \lim_{h \rightarrow 0 }\ -4x-2h+11 \]

the "limit" idea is let h approach 0 as close as we like... that makes -4x+11 -2h approach -4x+11

(so yes, the h "disappears")

Answer 24

What did you do to the -2?

Answer 25

Oh! Did you make h=0?

Answer 26

You can think of it as letting h=0, but to be technically correct we let h get unbelievably small , but not zero. We don't want h to be zero because we divided by h and dividing by 0 is "bozo no-no" so h is not zero. But it is so small we can think of it as zero (morally speaking)

Answer 27

Ahhh. So anything connected to the h (in this case, our -2) can be considered gone. That's why I kept getting -38. Sheesh. So it's -25. Yay!

Answer 28

This whole idea of infinitesimals was very controversial and confusing when Newton invented them. It took a long time for people to come up with this limit idea that makes the math work out.

Answer 29

Yeah, I can see why. Do you mind helping me with one more just to make sure Iv'e got it?

Answer 30

make it a new post.