Question

Is this correct? (Spoiler: It's derivatives)

http://i.imgur.com/MBKQYup.png

Answer 1

what is the derivative of
\[ 7 x^{-1} \] ?

Answer 2

using the power rule d/dx x^n = n x^(n-1)

Answer 3

I wish I could tell you - the content doesn't cover that. It only gives me f(x) and ask me to find it.

Answer 4

The rule is posted above. But here it is again
\[ \frac{d}{dx} x^n = n x^{n-1} \]

Answer 5

in your case, n= -1

Answer 6

What would variable "d" be then?

Answer 7

The \(d\) is notation for an operator. Just like \(+\), it doesn't have a value

Answer 8

d/dx is how you say "derivative with respect to x"

Answer 9

the power rule for finding a derivative is one of the first things you learn in differential calculus. If this is mysterious, you need a refresher, because it is too much material to review here.

Answer 10

I'm in Precalc, so we're probably never gonna cover the power rule. This is simply using the difference quotient.

Answer 11

you mean using
\[ f'(x) = \lim_{h\rightarrow0}\frac{ f(x+h) - f(x)}{(x+h)-x} \]?

Answer 12

Yep, we're doing that to solve for the posted equation.

Answer 13

in that case write
\[ f'(x) = \lim_{h\rightarrow0}\frac{\frac{7}{x+h} - \frac{7}{x} }{h} \]

Answer 14

Alright, sorry. Internet kinda cut out.
I know we gotta find a common denominator between the two fractions.

Answer 15

(which would be x for the first fraction, (x+h) for the second if I'm correct in saying that)

Answer 16

or
\[ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x} \right)\]

Answer 17

multiply the first fraction by x/x and the second by (x+h)/(x+h)

Answer 18

\[\frac{ x }{ x^2 +xh} - \frac{ x+h }{ x^2+xh }\] Right?

Answer 19

ok. now combine the top, and put the result over the common denominator

Answer 20

\[\frac{ h }{ x^2+xh }\] Alrightie.

Answer 21

the top is x-(x+h) = x-x-h

Answer 22

\[ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x} \right) \\
f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{-h}{x^2+xh} \right)
\]

Answer 23

Alright, I see.

Answer 24

notice h/h divides out and you have
\[ f'(x) = 7\lim_{h\rightarrow0} \left(\frac{-1}{x^2+xh} \right) \]

Answer 25

now let h "go to zero". the expression approaches -1/(x^2)
you get
\[ f'(x) = 7 \cdot \frac{-1}{x^2} \]
evaluate that at x=1

Answer 26

So from here I'd replace x^2 with 1?

Answer 27

you replace x with 1 so you get 1^2 or 1*1 in the denominator.

Answer 28

Which gives the end result of -7?

Answer 29

yes

Answer 30

Alright, thank you so much for your help.

Answer 31

yw