Answered: "Using the limit definition, how do you find..."

Porter Mccullough

Answered question

2022-04-19

Using the limit definition, how do you find the derivative of

f (x) = 2 \sqrt{x}

morpheus1ls1

Beginner2022-04-20Added 22 answers

Using the limit definition of a derivative, we have

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

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

= 2 lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}

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

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

= 2 \cdot \frac{1}{\sqrt{x + 0} + \sqrt{x}}

= 2 \cdot \frac{1}{2 \sqrt{x}}

= \frac{1}{\sqrt{x}}

mandamosppy

Beginner2022-04-21Added 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{2 \sqrt{x + h} - 2 \sqrt{x}}{h})

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

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