Derivative of 1/x

  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. why not?
     
  4. [tex](\frac{1}{x})'=(x^{-1})'=-x^{-1-1}=-\frac{1}{x^2}[/tex]
     
  5. 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. The tangent to the curve slopes downward. The negative sign is correct.
     
  7. 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. [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. Char. Limit

    Char. Limit 1,986
    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.
     
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