How do I use the limit definition of derivative to find f'(x) for f(x)=x+3?

By definition:ddx(x+3)=limh→0x+h+3−x+3hddx(x+3)=limh→0(x+h+3−x+3h)(x+h+3+x+3x+h+3+x+3)ddx(x+3)=limh→0(x+h+3)−(x+3)h(x+h+3+x+3)ddx(x+3)=limh→0hh(x+h+3+x+3)ddx(x+3)=limh→01x+h+3+x+3ddx(x+3)=12x+3

The derivative of f(x) with respect to x is defined as:limh→0[f(x+h)−f(x)h]=limh→0(x+h+3−x+3h)=limh→0(x+h+3−x+3h)=12x+3=12x+3

By definition:ddx(x+3)=limh→0x+h+3−x+3hddx(x+3)=limh→0(x+h+3−x+3h)(x+h+3+x+3x+h+3+x+3)ddx(x+3)=limh→0(x+h+3)−(x+3)h(x+h+3+x+3)ddx(x+3)=limh→0hh(x+h+3+x+3)ddx(x+3)=limh→01x+h+3+x+3ddx(x+3)=12x+3

The derivative of f(x) with respect to x is defined as:limh→0[f(x+h)−f(x)h]=limh→0(x+h+3−x+3h)=limh→0(x+h+3−x+3h)=12x+3=12x+3

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River Dudley

Answered question

2022-04-17

How do I use the limit definition of derivative to find f'(x) for

f (x) = \sqrt{x + 3}

Answer & Explanation

o2z1zsjj0

Beginner2022-04-18Added 10 answers

By definition:

\frac{d}{d x} (\sqrt{x + 3}) = lim_{h \to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h}

\frac{d}{d x} (\sqrt{x + 3}) = lim_{h \to 0} (\frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h}) (\frac{\sqrt{x + h + 3} + \sqrt{x + 3}}{\sqrt{x + h + 3} + \sqrt{x + 3}})

\frac{d}{d x} (\sqrt{x + 3}) = lim_{h \to 0} \frac{(x + h + 3) - (x + 3)}{h (\sqrt{x + h + 3} + \sqrt{x + 3})}

\frac{d}{d x} (\sqrt{x + 3}) = lim_{h \to 0} \frac{h}{h (\sqrt{x + h + 3} + \sqrt{x + 3})}

\frac{d}{d x} (\sqrt{x + 3}) = lim_{h \to 0} \frac{1}{\sqrt{x + h + 3} + \sqrt{x + 3}}

\frac{d}{d x} (\sqrt{x + 3}) = \frac{1}{2 \sqrt{x + 3}}

awalkbyfaithbzu6

Beginner2022-04-19Added 21 answers

The derivative of f(x) with respect to x is defined as:

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

= lim_{h \to 0} (\frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h})

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

= \frac{1}{2 \sqrt{x + 3}}

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