How do you find f'(x) using the definition of a derivative for f(x)=2x2−x?

f′(x)=limh→0f(x+h)−f(x)hlimh→02(x+h)2−(x+h)−(2x2−x)h=limh→02x2+4xh+2h2−x−h−2x2+xh=limh→04xh+2h2−hh=limh→0(4x+2h−1)=4x−1

differentiating from first principlesf′(x)=limh→0f(x+h)−f(x)h=limh→02(x+h)2−(x+h)−(2x2−x)h=limh→02x2+4hx+2h2−x−h−2x2+xh=limh→04hx+2h2−hh=limh→0h(4x+2h-1)h=4x-1

f′(x)=limh→0f(x+h)−f(x)hlimh→02(x+h)2−(x+h)−(2x2−x)h=limh→02x2+4xh+2h2−x−h−2x2+xh=limh→04xh+2h2−hh=limh→0(4x+2h−1)=4x−1

differentiating from first principlesf′(x)=limh→0f(x+h)−f(x)h=limh→02(x+h)2−(x+h)−(2x2−x)h=limh→02x2+4hx+2h2−x−h−2x2+xh=limh→04hx+2h2−hh=limh→0h(4x+2h-1)h=4x-1

Step by step solution to "How do you find f'(x) using the definition..."

High School
Calculus and Analysis
Calculus 1
Derivatives

painter555ui8n

Answered question

2022-04-15

How do you find f'(x) using the definition of a derivative for

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

Answer & Explanation

shulsellsmine2aaf

Beginner2022-04-16Added 5 answers

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

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

lim_{h \to 0} \frac{2 x^{2} + 4 x h + 2 h^{2} - x - h - 2 x^{2} + x}{h} =

lim_{h \to 0} \frac{4 x h + 2 h^{2} - h}{h} = lim_{h \to 0} (4 x + 2 h - 1) = 4 x - 1

gigglesbuggk1co

Beginner2022-04-17Added 13 answers

differentiating from first principles
$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
$= lim_{h \to 0} \frac{2 {(x + h)}^{2} - (x + h) - (2 x^{2} - x)}{h}$
$= lim_{h \to 0} \frac{2 x^{2} + 4 h x + 2 h^{2} - x - h - 2 x^{2} + x}{h}$
$= lim_{h \to 0} \frac{4 h x + 2 h^{2} - h}{h}$
$= \lim_{h \to 0} \frac{h (4 x + 2 h - 1)}{h}$
=4x-1

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