How do you find the derivative using limits of f(x)=1x2?

The limit definition of the derivative of a function f(x) is:f′(x)=lim△x→0f(x+△x)−f(x)△x=lim△x→0△f△xLet's calculate the increment of the function between x and x+△x:△f=1(x+△x)2−1x2=x2−(x+△x)2x2(x+△x)2=x2−x2−2x△x−(△x)2x2(x+△x)2=−2x△x−(△x)2x2(x+△x)2The incremental ratio is then:△f△x=−2x−△xx2(x+△x)2and passing to the limit:lim△x→0−2x−△xx2(x+△x)2=−2xx2⋅x2=−2x3

Using the limit definition for a derivative:f′(x)=limh→0f(x+h)−f(x)hFor the case f(x)=1x2f(x+h)−f(x)=1(x+h)2−1x2=x2−(x+h)2x2⋅(x+h)2=−2xh−h2x4+2hx3+h2x2and thereforef(x+h)−f(x)h=−2x−hx4+2hx3+h2x2This is defined when h=0solimh→0f(x+h)−f(x)h=−2x−0x4+2⋅0⋅x3+02⋅x2=−2xx4=−2x3=(−2)⋅(1x3)

The limit definition of the derivative of a function f(x) is:f′(x)=lim△x→0f(x+△x)−f(x)△x=lim△x→0△f△xLet's calculate the increment of the function between x and x+△x:△f=1(x+△x)2−1x2=x2−(x+△x)2x2(x+△x)2=x2−x2−2x△x−(△x)2x2(x+△x)2=−2x△x−(△x)2x2(x+△x)2The incremental ratio is then:△f△x=−2x−△xx2(x+△x)2and passing to the limit:lim△x→0−2x−△xx2(x+△x)2=−2xx2⋅x2=−2x3

Using the limit definition for a derivative:f′(x)=limh→0f(x+h)−f(x)hFor the case f(x)=1x2f(x+h)−f(x)=1(x+h)2−1x2=x2−(x+h)2x2⋅(x+h)2=−2xh−h2x4+2hx3+h2x2and thereforef(x+h)−f(x)h=−2x−hx4+2hx3+h2x2This is defined when h=0solimh→0f(x+h)−f(x)h=−2x−0x4+2⋅0⋅x3+02⋅x2=−2xx4=−2x3=(−2)⋅(1x3)

Answered: "How do you find the derivative using limits..."

Reagan Gomez

Answered question

2022-04-15

How do you find the derivative using limits of

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

Answer & Explanation

Kaphefsceasiaf8w9

Beginner2022-04-16Added 9 answers

The limit definition of the derivative of a function f(x) is:

f^{'} (x) = lim_{△ x \to 0} \frac{f (x + △ x) - f (x)}{△ x} = lim_{△ x \to 0} \frac{△ f}{△ x}

Let's calculate the increment of the function between x and

x + △ x :

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

= \frac{- 2 x △ x - {(△ x)}^{2}}{x^{2} {(x + △ x)}^{2}}

The incremental ratio is then:

\frac{△ f}{△ x} = \frac{- 2 x - △ x}{x^{2} {(x + △ x)}^{2}}

and passing to the limit:

lim_{△ x \to 0} \frac{- 2 x - △ x}{x^{2} {(x + △ x)}^{2}} = - 2 \frac{x}{x^{2} \cdot x^{2}} = - \frac{2}{x^{3}}

Dallelopeelvep2yc

Beginner2022-04-17Added 15 answers

Using the limit definition for a derivative:

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

For the case

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

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

= \frac{x^{2} - {(x + h)}^{2}}{x^{2} \cdot {(x + h)}^{2}}

= \frac{- 2 x h - h^{2}}{x^{4} + 2 h x^{3} + h^{2} x^{2}}

and therefore

\frac{f (x + h) - f (x)}{h} = \frac{- 2 x - h}{x^{4} + 2 h x^{3} + h^{2} x^{2}}

This is defined when h=0
so

lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \frac{- 2 x - 0}{x^{4} + 2 \cdot 0 \cdot x^{3} + 0^{2} \cdot x^{2}}

= \frac{- 2 x}{x^{4}} = - \frac{2}{x^{3}} = (- 2) \cdot (\frac{1}{x^{3}})

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