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 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]
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
[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? :tongue:
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.