How do you find f'(x) using the limit definition given f(x)=3x−2?

Explanation:by definitionf′(x)=limh→0f(x+h)−f(x)h=limh→0(1h)(3(x+h)2−3x2)=limh→0(1h)⋅3x2−3(x+h)2(x+h)2x2=limh→0(1h)⋅3x2−3(x2+2xh+h2)(x+h)2x2=limh→0(1h)⋅−6xh−3h2(x+h)2x2=limh→0−6x−3h(x+h)2x2=limh→0−6x(x+h)2x2=−6x(x+0)2x2=−6xx4=−6x3

Using the limit definition we get:f′(x)=limh→0f(x+h)−f(x)hwhere h is a small increment.In our case we have (using the fact that&nbsp;x−2=1x2):f′(x)=limh→03(x+h)2−3x2h=limh→01h(3x2−3(x+h)2(x+h)2x2)=limh→01h(3x2−3x2−6xh−3h2(x+h)2x2)==limh→01h(h-6x-3h(x+h)2x2)=as&nbsp;h→0&nbsp;we get:=limh→0(−6x−3h0(x+h0)2x2)=−6xx4=−6x3

Explanation:by definitionf′(x)=limh→0f(x+h)−f(x)h=limh→0(1h)(3(x+h)2−3x2)=limh→0(1h)⋅3x2−3(x+h)2(x+h)2x2=limh→0(1h)⋅3x2−3(x2+2xh+h2)(x+h)2x2=limh→0(1h)⋅−6xh−3h2(x+h)2x2=limh→0−6x−3h(x+h)2x2=limh→0−6x(x+h)2x2=−6x(x+0)2x2=−6xx4=−6x3

Using the limit definition we get:f′(x)=limh→0f(x+h)−f(x)hwhere h is a small increment.In our case we have (using the fact that&nbsp;x−2=1x2):f′(x)=limh→03(x+h)2−3x2h=limh→01h(3x2−3(x+h)2(x+h)2x2)=limh→01h(3x2−3x2−6xh−3h2(x+h)2x2)==limh→01h(h-6x-3h(x+h)2x2)=as&nbsp;h→0&nbsp;we get:=limh→0(−6x−3h0(x+h0)2x2)=−6xx4=−6x3

Find out the answer to "How do you find f'(x) using the limit..."

Bradley Barron

Answered question

2022-04-14

How do you find f'(x) using the limit definition given

f (x) = 3 x^{- 2}

Answer & Explanation

Vegljamzt6

Beginner2022-04-15Added 16 answers

Explanation:
by definition

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

= lim_{h \to 0} (\frac{1}{h}) (\frac{3}{{(x + h)}^{2}} - \frac{3}{x^{2}})

= lim_{h \to 0} (\frac{1}{h}) \cdot \frac{3 x^{2} - 3 {(x + h)}^{2}}{{(x + h)}^{2} x^{2}}

= lim_{h \to 0} (\frac{1}{h}) \cdot \frac{3 x^{2} - 3 (x^{2} + 2 x h + h^{2})}{{(x + h)}^{2} x^{2}}

= lim_{h \to 0} (\frac{1}{h}) \cdot \frac{- 6 x h - 3 h^{2}}{{(x + h)}^{2} x^{2}}

= lim_{h \to 0} \frac{- 6 x - 3 h}{{(x + h)}^{2} x^{2}}

= lim_{h \to 0} \frac{- 6 x}{{(x + h)}^{2} x^{2}}

= - \frac{6 x}{{(x + 0)}^{2} x^{2}}

= - \frac{6 x}{x^{4}}

= - \frac{6}{x^{3}}

Buizzae77t

Beginner2022-04-16Added 13 answers

Using the limit definition we get:
$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
where h is a small increment.
In our case we have (using the fact that $x^{- 2} = \frac{1}{x^{2}}$ ):
$f^{'} (x) = lim_{h \to 0} \frac{\frac{3}{{(x + h)}^{2}} - \frac{3}{x^{2}}}{h}$
$= lim_{h \to 0} \frac{1}{h} (\frac{3 x^{2} - 3 {(x + h)}^{2}}{{(x + h)}^{2} x^{2}})$
$= lim_{h \to 0} \frac{1}{h} (\frac{3 x^{2} - 3 x^{2} - 6 x h - 3 h^{2}}{{(x + h)}^{2} x^{2}}) =$
$= \lim_{h \to 0} \frac{1}{h} (h \frac{- 6 x - 3 h}{(x + h)^{2} x^{2}}) =$
as $h \to 0$ we get:
$= lim_{h \to 0} (\frac{- 6 x - 3 h^{0}}{{(x + h^{0})}^{2} x^{2}}) = \frac{- 6 x}{x^{4}} = - \frac{6}{x^{3}}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?