# Derivative of 1/x

by gabby989062
Tags: 1 or x, derivative
 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?
P: 2,251
 Quote by 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?

why not?
 P: 38 $$(\frac{1}{x})'=(x^{-1})'=-x^{-1-1}=-\frac{1}{x^2}$$
 P: 8 Derivative of 1/x Why does it become negative at this jump? $$(\frac{1}{x})'=(x^{-1})'=-x^{-1-1}=-\frac{1}{x^2}$$ Why not $$\frac{1}{x^2}$$
 P: 1,781 The tangent to the curve slopes downward. The negative sign is correct.
 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
 P: 534 $$f(x) \ = \ \frac{1}{x} \ = \ \frac{1}{x^1} \ = \ x^{(-1)}$$ $$f'(x) \ = \ (-1)x^{(-1) - 1} \ = \ (-1)x^{(-2)} \ = \ \frac{(-1)}{x^{2}} \ =\ - \ \frac{1}{x^2}$$ 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?
PF Gold
P: 1,951
 Quote by sponsoredwalk $$f(x) \ = \ \frac{1}{x} \ = \ \frac{1}{x^1} \ = \ x^{(-1)}$$ $$f'(x) \ = \ (-1)x^{(-1) - 1} \ = \ (-1)x^{(-2)} \ = \ \frac{(-1)}{x^{2}} \ =\ - \ \frac{1}{x^2}$$ 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?
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.

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