anonymous
  • anonymous
Is this correct? (Spoiler: It's derivatives) http://i.imgur.com/MBKQYup.png
Mathematics
katieb
  • katieb
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phi
  • phi
what is the derivative of \[ 7 x^{-1} \] ?
phi
  • phi
using the power rule d/dx x^n = n x^(n-1)
anonymous
  • anonymous
I wish I could tell you - the content doesn't cover that. It only gives me f(x) and ask me to find it.

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phi
  • phi
The rule is posted above. But here it is again \[ \frac{d}{dx} x^n = n x^{n-1} \]
phi
  • phi
in your case, n= -1
anonymous
  • anonymous
What would variable "d" be then?
anonymous
  • anonymous
The \(d\) is notation for an operator. Just like \(+\), it doesn't have a value
phi
  • phi
d/dx is how you say "derivative with respect to x"
phi
  • phi
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.
anonymous
  • anonymous
I'm in Precalc, so we're probably never gonna cover the power rule. This is simply using the difference quotient.
phi
  • phi
you mean using \[ f'(x) = \lim_{h\rightarrow0}\frac{ f(x+h) - f(x)}{(x+h)-x} \]?
anonymous
  • anonymous
Yep, we're doing that to solve for the posted equation.
phi
  • phi
in that case write \[ f'(x) = \lim_{h\rightarrow0}\frac{\frac{7}{x+h} - \frac{7}{x} }{h} \]
anonymous
  • anonymous
Alright, sorry. Internet kinda cut out. I know we gotta find a common denominator between the two fractions.
anonymous
  • anonymous
(which would be x for the first fraction, (x+h) for the second if I'm correct in saying that)
phi
  • phi
or \[ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x} \right)\]
phi
  • phi
multiply the first fraction by x/x and the second by (x+h)/(x+h)
anonymous
  • anonymous
\[\frac{ x }{ x^2 +xh} - \frac{ x+h }{ x^2+xh }\] Right?
phi
  • phi
ok. now combine the top, and put the result over the common denominator
anonymous
  • anonymous
\[\frac{ h }{ x^2+xh }\] Alrightie.
phi
  • phi
the top is x-(x+h) = x-x-h
phi
  • phi
\[ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x} \right) \\ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{-h}{x^2+xh} \right) \]
anonymous
  • anonymous
Alright, I see.
phi
  • phi
notice h/h divides out and you have \[ f'(x) = 7\lim_{h\rightarrow0} \left(\frac{-1}{x^2+xh} \right) \]
phi
  • phi
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
anonymous
  • anonymous
So from here I'd replace x^2 with 1?
phi
  • phi
you replace x with 1 so you get 1^2 or 1*1 in the denominator.
anonymous
  • anonymous
Which gives the end result of -7?
phi
  • phi
yes
anonymous
  • anonymous
Alright, thank you so much for your help.
phi
  • phi
yw

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