Question

Find the derivative of f(x) = 6/x at x = -2.

Answer 1

solomon help haha

Answer 2

Ok, lets re-write the f(x)

\(\large\color{black}{ \displaystyle f(x)=6(x)^{-1} }\)

Answer 3

Apply the power rule.
Can you do that?

Answer 4

(YOu are to find the derivative, and then plug in x=-2 into the derivative)

Answer 5

-3

Answer 6

is that your fnal answer?

Answer 7

if so, then you are not correct....

Answer 8

Did you find the \(f'(x)\) /?

Answer 9

Oh, what I mean by the power rule is:

\(\large\color{black}{ \displaystyle \frac{d }{dx}x^n=nx^{n-1} }\)

where d/dx is jst a notation for taking the derivative.
------------------------------------------------
But I guess you are doing by the first principles...

Answer 10

'm not really sure what you mean by power rule I have a formula for difference quotient
f(h-1)-f(1)/h and I ended up with ([6/h-1]-6)/h

Answer 11

never heard of it this is precalc

Answer 12

yes, you are applying the following:

\(\large\color{black}{ \displaystyle \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)

Answer 13

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)

Answer 14

YOu can use the power rule I posted, to at least check the work, but for now I guess we need this:

\(\large\color{black}{ \displaystyle \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)

Answer 15

Or, if you want to find \(f'(a)\) direclty:

\(\large\color{black}{ \displaystyle f'(a)= \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} }\)

Answer 16

(6/x+h)-(6/x)/h

Answer 17

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6}{x+h}-\dfrac{6}{x} }{h} }\)

Answer 18

that is right.

Answer 19

now, find the common denominator betwen 6/(x+h) and 6/x
and subtract.

Answer 20

so would I multiply one side by x and the other by x+h

Answer 21

yes, fraction#1 by x on top and bottom, and fraction#2 (x+h) on top and bottom

Answer 22

I got -6

Answer 23

pretty sure that's it thanks

Answer 24

no it is not it

Answer 25

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6}{x+h}-\dfrac{6}{x} }{h} }\)

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6x}{x(x+h)}-\dfrac{6(x+h)}{x(x+h)} }{h} }\)

Answer 26

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6x-6(x+h)}{x(x+h)} }{h} }\)

Answer 27

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{-6h}{x(x+h)} }{h} }\)

Answer 28

yep that's what I got just forgot about the denomenator under the -6h

Answer 29

then divide top and bottom by h.

Answer 30

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{-6h}{x(x+h)}\color{red}{\div h} }{h \color{red}{\div h}} }\)

Answer 31

then you get:

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{-6}{x(x+h)} }\)

Answer 32

-6/x(x+h)

Answer 33

yes, but you are leaving out that important limit.

Answer 34

so I just put -2 in for x

Answer 35

that limit that h=0, is an important component.
So that when you simplify the expression, you then plug in h=0
(if you don;t get any undefined results for that)

Answer 36

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{-6}{x(x+h)} =\frac{-6}{x(x+0)}=\frac{-6}{x^2} }\)

Answer 37

see what is that limit for?
(it is a notation for the fact that h is 0)

Answer 38

- 3/2

Answer 39

yes, that is right:

Answer 40

hey, thanks for the patience I'm kind of slow

Answer 41

Yes, don't forget that limit h->0 notation. it is important.

Answer 42

So just an addition that in general:

\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{f(x+h)-f(x)}{h} }\)
(Derivative a function f(x).)


\(\large\color{black}{ \displaystyle f'(x)= \lim_{x \rightarrow a}{~} \frac{f(x)-f(a)}{x-a} }\)
(Derivative a function f(x) evaluated at x=a.)

Answer 43

good luck