# Derivative of 1/x

1. Jan 28, 2008

### gabby989062

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. Jan 28, 2008

### rbj

why not?

3. Jan 28, 2008

### fermio

$$(\frac{1}{x})'=(x^{-1})'=-x^{-1-1}=-\frac{1}{x^2}$$

4. Apr 23, 2010

### literacola

Why does it become negative at this jump?

$$(\frac{1}{x})'=(x^{-1})'=-x^{-1-1}=-\frac{1}{x^2}$$

Why not $$\frac{1}{x^2}$$

5. Apr 23, 2010

### Antiphon

The tangent to the curve slopes downward. The negative sign is correct.

6. Aug 19, 2010

### ykalson

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

7. Aug 19, 2010

$$f(x) \ = \ \frac{1}{x} \ = \ \frac{1}{x^1} \ = \ x^{(-1)}$$

$$f'(x) \ = \ (-1)x^{(-1) - 1} \ = \ (-1)x^{(-2)} \ = \ \frac{(-1)}{x^{2}} \ =\ - \ \frac{1}{x^2}$$

Proof?

$$f(x) \ = \ \frac{1}{x}$$

$$f(x \ + \ h) \ = \ \frac{1}{x + h}$$

$$f(x \ + \ h) \ - \ f(x) \ = \ \frac{1}{x + h} \ - \ \frac{1}{x}$$

$$f(x \ + \ h) \ - \ f(x) \ = \ \frac{1}{x + h} \ - \ \frac{1}{x} \ \Rightarrow \ \frac{x \ - \ x \ - \ h }{x(x + h)}$$

$$\lim_{h \to 0} \ \frac{f(x \ + \ h) \ - \ f(x)}{h} \ = \ ?$$

Can you finish it off person who originally asked this question over 2 years ago? :tongue:

8. Aug 20, 2010

### Char. Limit

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

IF

$$f(x+h) - f(x) = \frac{x-x-h}{x(x+h)}$$

THEN

$$f(x+h) - f(x) = \frac{-h}{x(x+h)}$$

THEN

$$\frac{f(x+h)-f(x)}{h} = \frac{-1}{x^2 + h x}$$

THEN

$$lim_{h\rightarrow0} \frac{f(x+h) - f(x)}{h} = \frac{-1}{x^2 + 0x} = \frac{-1}{x^2} = f'(x)$$

Quod Erat Demonstrandum.