
#1
Jan2808, 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? 



#2
Jan2808, 01:00 AM

P: 2,265

why not? 



#3
Jan2808, 01:13 AM

P: 38

[tex](\frac{1}{x})'=(x^{1})'=x^{11}=\frac{1}{x^2}[/tex]




#4
Apr2310, 11:01 PM

P: 8

derivative of 1/x
Why does it become negative at this jump?
[tex](\frac{1}{x})'=(x^{1})'=x^{11}=\frac{1}{x^2}[/tex] Why not [tex]\frac{1}{x^2}[/tex] 



#5
Apr2310, 11:05 PM

P: 1,781

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




#6
Aug1910, 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 



#7
Aug1910, 09:37 PM

P: 530

[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? 



#8
Aug2010, 12:46 AM

PF Gold
P: 1,930

IF [tex] f(x+h)  f(x) = \frac{xxh}{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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