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Derivative (examples)

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Example 1

Consider f(x) = 5:

$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0} \frac{5-5}{h} = 0$

The derivative of a constant is zero.

Example 2

Consider the graph of $f (x) = 2 x - 3$ . If the reader has an understanding of algebra and the Cartesian coordinate system, the reader should be able to independently determine that this line has a slope of 2 at every point. Using the above quotient (along with an understanding of the limit, secant, and tangent) one can determine the slope at (4,5):

$f'(4) = \lim_{h\rightarrow 0}\frac{f(4+h)-f(4)}{h}$

$= \lim_{h\rightarrow 0}\frac{2(4+h)-3-(2\cdot 4-3)}{h}$

$= \lim_{h\rightarrow 0}\frac{8+2h-3-8+3}{h}$

$= \lim_{h\rightarrow 0}\frac{2h}{h}$

= 2

The derivative and slope are equivalent.

Example 3

Via differentiation, one can find the slope of a curve. Consider $f (x) = x 2$ :

$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

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

$= \lim_{h\rightarrow 0}\frac{x^2 + 2xh + h^2 - x^2}{h}$

$= \lim_{h\rightarrow 0}\frac{2xh + h^2}{h}$

$= \lim_{h\rightarrow 0}(2x + h)$

= 2 x

For any point x, the slope of the function $f (x) = x 2$ is $f'(x) = 2 x$ .

Example 4

Consider f(x) = √x:

$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$= \lim_{h\rightarrow 0}\frac{\sqrt{x+h} - \sqrt{x}}{h}$

$= \lim_{h\rightarrow 0}\frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}$

$= \lim_{h\rightarrow 0}\frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}$

$= \lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h} + \sqrt{x}}$

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

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