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

limh→0[(x+h)2+x+h]−[x2+x]hlimh→0[x2+2xh+h2+x+h]−[x2+x]hlimh→0x2+2xh+h2+x+h−x2−xhlimh→02xh+h2+hhlimh→0h(2x+h+1)hlimh→02x+h+1=2x+0+1=2x+1Answer:f'(x)=2x+1

limh→0f(x+h)−f(x)h=limh→0(x+h)2+(x+h)−x2−xh=limh→0x2+2xh+h2+h−x2h=limh→0h(2x+h+1)h=limh→0(2x+1+h)=2x+1.

limh→0[(x+h)2+x+h]−[x2+x]hlimh→0[x2+2xh+h2+x+h]−[x2+x]hlimh→0x2+2xh+h2+x+h−x2−xhlimh→02xh+h2+hhlimh→0h(2x+h+1)hlimh→02x+h+1=2x+0+1=2x+1Answer:f'(x)=2x+1

limh→0f(x+h)−f(x)h=limh→0(x+h)2+(x+h)−x2−xh=limh→0x2+2xh+h2+h−x2h=limh→0h(2x+h+1)h=limh→0(2x+1+h)=2x+1.

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High School
Calculus and Analysis
Calculus 1
Derivatives

Aliyah Mendez

Answered question

2022-04-18

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

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

Answer & Explanation

gigglesbuggk1co

Beginner2022-04-19Added 13 answers

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

lim_{h \to 0} \frac{[x^{2} + 2 x h + h^{2} + x + h] - [x^{2} + x]}{h}

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

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

lim_{h \to 0} \frac{h (2 x + h + 1)}{h}

lim_{h \to 0} 2 x + h + 1 = 2 x + 0 + 1 = 2 x + 1

Answer:
f'(x)=2x+1

szalbierzfytg

Beginner2022-04-20Added 13 answers

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

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

= lim_{h \to 0} (2 x + 1 + h) = 2 x + 1

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