1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative of 1/x

  1. Jan 28, 2008 #1
    1. The problem statement, all variables and given/known data
    This is one step of a larger problem, but I'm stuck on derivative of 1/x.


    2. Relevant equations



    3. The attempt at a solution
    1/x = x^-1.

    Using power rule:
    -x^-2
    but this isn't right?
     
  2. jcsd
  3. Jan 28, 2008 #2

    rbj

    User Avatar


    why not?
     
  4. Jan 28, 2008 #3
    [tex](\frac{1}{x})'=(x^{-1})'=-x^{-1-1}=-\frac{1}{x^2}[/tex]
     
  5. Apr 23, 2010 #4
    Why does it become negative at this jump?

    [tex](\frac{1}{x})'=(x^{-1})'=-x^{-1-1}=-\frac{1}{x^2}[/tex]

    Why not [tex]\frac{1}{x^2}[/tex]
     
  6. Apr 23, 2010 #5
    The tangent to the curve slopes downward. The negative sign is correct.
     
  7. Aug 19, 2010 #6
    You can also use the Quotient Rule...

    Which says that (f(x)/g(x))' = [f'(x)g(x)-f(x)g'(x)]/(g(x)^2)

    So we let f=1 and g=x

    and we compute that f'=0 and g'=1

    Then just put it all together... [(0)(x)-(1)(1)]/(x^2)

    This gives us -1/x^2
     
  8. Aug 19, 2010 #7
    [tex] f(x) \ = \ \frac{1}{x} \ = \ \frac{1}{x^1} \ = \ x^{(-1)}[/tex]

    [tex] f'(x) \ = \ (-1)x^{(-1) - 1} \ = \ (-1)x^{(-2)} \ = \ \frac{(-1)}{x^{2}} \ =\ - \ \frac{1}{x^2} [/tex]

    Proof?

    [tex] f(x) \ = \ \frac{1}{x} [/tex]

    [tex] f(x \ + \ h) \ = \ \frac{1}{x + h}[/tex]

    [tex] f(x \ + \ h) \ - \ f(x) \ = \ \frac{1}{x + h} \ - \ \frac{1}{x} [/tex]

    [tex] f(x \ + \ h) \ - \ f(x) \ = \ \frac{1}{x + h} \ - \ \frac{1}{x} \ \Rightarrow \ \frac{x \ - \ x \ - \ h }{x(x + h)} [/tex]

    [tex] \lim_{h \to 0} \ \frac{f(x \ + \ h) \ - \ f(x)}{h} \ = \ ? [/tex]

    Can you finish it off person who originally asked this question over 2 years ago? :tongue:
     
  9. Aug 20, 2010 #8

    Char. Limit

    User Avatar
    Gold Member

    Since he's likely gone, I'll finish it for him, for the benefit of anyone confused to look at this page.

    IF

    [tex] f(x+h) - f(x) = \frac{x-x-h}{x(x+h)}[/tex]

    THEN

    [tex] f(x+h) - f(x) = \frac{-h}{x(x+h)}[/tex]

    THEN

    [tex]\frac{f(x+h)-f(x)}{h} = \frac{-1}{x^2 + h x}[/tex]

    THEN

    [tex]lim_{h\rightarrow0} \frac{f(x+h) - f(x)}{h} = \frac{-1}{x^2 + 0x} = \frac{-1}{x^2} = f'(x)[/tex]

    Quod Erat Demonstrandum.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Derivative of 1/x
  1. Derivative of 1 / ln x (Replies: 6)

  2. Derivative 1/x³ (Replies: 5)

Loading...