Derivative (examples)

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For more background on this topic, see derivative.

1 Example 1
2 Example 2
3 Example 3
4 Example 4
5 Example 5

[edit] Example 1

Consider f(x) = 5:

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

The derivative of a constant function is zero.

[edit] 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):

$\begin{align} f'(4) &= \lim_{h\to 0}\frac{f(4+h)-f(4)}{h} \\ &= \lim_{h\to 0}\frac{2(4+h)-3-(2\cdot 4-3)}{h} \\ &= \lim_{h\to 0}\frac{8+2h-3-8+3}{h} \\ &= \lim_{h\to 0}\frac{2h}{h} \\ &= 2 \end{align}$

The derivative and slope are equivalent.

[edit] Example 3

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

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

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

[edit] Example 4

Consider $f(x) = \sqrt{x}$ :

$f'(x)\,$	$= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
	$= \lim_{h\rightarrow 0}\frac{\sqrt{x+h} - \sqrt{x}}{h}$
	$= \lim_{h\rightarrow 0}\left(\frac{\sqrt{x+h} - \sqrt{x}}{h}\right) \left(\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\right)$
	$= \lim_{h\rightarrow 0}\frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}$
	$= \lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h} + \sqrt{x}}$
	$= \frac{1 }{2 \sqrt{x}}$

[edit] Example 5

The same as the previous example, but now we search the derivative of the derivative.
Consider $f(x) = \sqrt{x}$ :

$\begin{align} f''(x) &= \lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h} \\ &= \lim_{h\to 0} \frac{\frac{1}{2 \sqrt{x+h}}-\frac{1}{2 \sqrt{x}}}{h} \\ &= \lim_{h\to 0} \frac{1}{2h}\left(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right) \\ &= \lim_{h\to 0} \frac{1}{2h}\left(\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}} \right) \\ &= \lim_{h\to 0} \frac{1}{2h}\left(\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}\right) \\ &= \lim_{h\to 0} \frac{1}{2h}\left(\frac{x - (x+h)}{x\sqrt{x+h} + (x+h)\sqrt{x} } \right) \\ &= \lim_{h\to 0} \frac{1}{2}\left(\frac{-1}{x\sqrt{x+h} + (x+h)\sqrt{x} } \right) \\ &= \frac{1}{2}\left(\frac{-1}{x\sqrt{x} + x\sqrt{x} } \right) \\ &= -\frac{1}{4 x \sqrt{x}} \end{align}$