# Derivative of 1/x the old fashioned way

1. Jun 16, 2011

### 1MileCrash

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]

ETC.

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

What would you guys do to algebraically solve this??

2. Jun 16, 2011

### micromass

Staff Emeritus
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).

3. Jun 16, 2011

### SteamKing

Staff Emeritus
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.

4. Jun 16, 2011

### 1MileCrash

That gets me no where.

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

or

-h / (x^2+hx)

5. Jun 16, 2011

### micromass

Staff Emeritus
Yes it will

And now you still need to divide through h...

6. Jun 16, 2011

### 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.

7. Jun 16, 2011

### 1MileCrash

That's where I'm stuck.

How can I get rid of the complex fraction here?

"Dividing through h" doesn't seem possible.

8. Jun 16, 2011

### Staff: Mentor

$$\lim_{h\rightarrow 0}{\frac{\frac{1}{x+h}-\frac{1}{x}}{h}} = \lim_{h \to 0} \frac{-h}{x^2 + hx}\cdot \frac{1}{h}$$

Can you simplify a bit more and then take the limit?

9. Jun 16, 2011

### 1MileCrash

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. Jun 16, 2011

Right!