Derivative of 1/x the old fashioned way

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SUMMARY

The derivative of the function f(x) = 1/x is calculated using the limit definition of the derivative, specifically f'(x) = lim h->0 [f(x+h) - f(x)] / [h]. Through algebraic manipulation, the expression simplifies to -1/(x^2) as h approaches 0. Key steps include correctly simplifying the difference of fractions and dividing through by h, leading to the final result of -1/x^2.

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Homework Statement



f(x) = 1/x

Find the derivative

Homework Equations



f'(x)= lim h->0

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

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]

ETC.

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

What would you guys do to algebraically solve this??
 
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You must calculate

\lim_{h\rightarrow 0}{\frac{\frac{1}{x+h}-\frac{1}{x}}{h}}

First simplify it (to the point where there's only one fraction, instead of three fractions).
 
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.
 
SteamKing said:
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.

That gets me no where.

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

or

-h / (x^2+hx)
 
1MileCrash said:
That gets me no where.

Yes it will :smile:

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

or

-h / (x^2+hx)

And now you still need to divide through h...
 
1MileCrash said:
That gets me no where.

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

or

-h / (x^2+hx)
This is not the complete quotient. This is just the simplification of the numerator, 1/x - 1/(x + h)

micromass said:
And now you still need to divide through h...
Dividing by h is equivalent to multiplying by 1/h.
 
That's where I'm stuck.

How can I get rid of the complex fraction here?

"Dividing through h" doesn't seem possible.
 
\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}

Can you simplify a bit more and then take the limit?
 
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
 
  • #10
1MileCrash said:
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

Right!
 

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