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Derivative of 1/x

by gabby989062
Tags: 1 or x, derivative
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gabby989062
#1
Jan28-08, 12:30 AM
P: 20
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?
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rbj
#2
Jan28-08, 01:00 AM
P: 2,251
Quote Quote by gabby989062 View Post
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?

why not?
fermio
#3
Jan28-08, 01:13 AM
P: 38
[tex](\frac{1}{x})'=(x^{-1})'=-x^{-1-1}=-\frac{1}{x^2}[/tex]

literacola
#4
Apr23-10, 11:01 PM
P: 8
Derivative of 1/x

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]
Antiphon
#5
Apr23-10, 11:05 PM
P: 1,781
The tangent to the curve slopes downward. The negative sign is correct.
ykalson
#6
Aug19-10, 08:28 PM
P: 1
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
sponsoredwalk
#7
Aug19-10, 09:37 PM
P: 534
[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?
Char. Limit
#8
Aug20-10, 12:46 AM
PF Gold
Char. Limit's Avatar
P: 1,951
Quote Quote by sponsoredwalk View Post
[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?
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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