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Derivative of 1/x the old fashioned way

  1. Jun 16, 2011 #1
    1. The problem statement, all variables and given/known data

    f(x) = 1/x

    Find the derivative

    2. Relevant equations

    f'(x)= lim h->0

    [f(x+h) - f(x)] / [h]

    3. The attempt at a solution

    I MUST show work and use that formula to arrive at the derivative (1/x^2)

    I can't figure out the algebra.

    I rearrange it countless ways..

    [1/(x+h) - 1/x] - (1/x)] / [h]


    But I can NEVER find a way to algebraically arrive at the derivative! Ever!

    What would you guys do to algebraically solve this??
  2. jcsd
  3. Jun 16, 2011 #2
    You must calculate

    [tex]\lim_{h\rightarrow 0}{\frac{\frac{1}{x+h}-\frac{1}{x}}{h}}[/tex]

    First simplify it (to the point where there's only one fraction, instead of three fractions).
  4. Jun 16, 2011 #3


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    Staff Emeritus
    Science Advisor
    Homework Helper

    Rearranging your limit expression is not necessarily going to help unless you can do the subtraction [1/(x+h) - 1/x] correctly. Show what you get for this calculation.
  5. Jun 16, 2011 #4
    That gets me no where.



    -h / (x^2+hx)
  6. Jun 16, 2011 #5
    Yes it will :smile:

    And now you still need to divide through h...
  7. Jun 16, 2011 #6


    Staff: Mentor

    This is not the complete quotient. This is just the simplification of the numerator, 1/x - 1/(x + h)

    Dividing by h is equivalent to multiplying by 1/h.
  8. Jun 16, 2011 #7
    That's where I'm stuck.

    How can I get rid of the complex fraction here?

    "Dividing through h" doesn't seem possible.
  9. Jun 16, 2011 #8


    Staff: Mentor

    [tex]\lim_{h\rightarrow 0}{\frac{\frac{1}{x+h}-\frac{1}{x}}{h}} = \lim_{h \to 0} \frac{-h}{x^2 + hx}\cdot \frac{1}{h}[/tex]

    Can you simplify a bit more and then take the limit?
  10. Jun 16, 2011 #9
    NOW I see! Thanks!

    -h / hx^2 + h^2x


    h(-1) / h(x^2+hx)=

    (-1) / x^2 + hx

    as h approaches 0, hx approaches zero, leaving me with x^2 in the denomenator for a final result of

    - 1/x^2
  11. Jun 16, 2011 #10


    Staff: Mentor

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