Derivative (examples)
- For more background on this topic, see derivative.
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.
Example 2
Consider the graph of f(x) = 2x − 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.
Example 3
Via differentiation, one can find the slope of a curve. Consider f(x) = x2:
- {|
|- |f′(x) |$= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ |- | |$= \lim_{h\rightarrow 0}\frac{(x+h)^2 - x^2}{h}$ |- | |$= \lim_{h\rightarrow 0}\frac{x^2 + 2xh + h^2 - x^2}{h}$ |- | |$= \lim_{h\rightarrow 0}\frac{2xh + h^2}{h}$ |- | | = limh → 0(2x+h) = 2x |}
For any point x, the slope of the function f(x) = x2 is f′(x) = 2x.
Example 4
Consider $f(x) = \sqrt{x}$:
- {|
|- |f′(x) |$= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ |- | |$= \lim_{h\rightarrow 0}\frac{\sqrt{x+h} - \sqrt{x}}{h}$ |- | |$= \lim_{h\rightarrow 0}\left(\frac{\sqrt{x+h} - \sqrt{x}}{h}\right) \left(\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\right)$ |- | |$= \lim_{h\rightarrow 0}\frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}$ |- | |$= \lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h} + \sqrt{x}}$ |- | |$= \frac{1 }{2 \sqrt{x}}$ |}
Example 5
The same as the previous example, but now we search the derivative of the derivative.
Consider $f(x) = \sqrt{x}$:
$$\begin{align} f''(x) &= \lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h} \\ &= \lim_{h\to 0} \frac{\frac{1}{2 \sqrt{x+h}}-\frac{1}{2 \sqrt{x}}}{h} \\ &= \lim_{h\to 0} \frac{1}{2h}\left(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right) \\ &= \lim_{h\to 0} \frac{1}{2h}\left(\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}} \right) \\ &= \lim_{h\to 0} \frac{1}{2h}\left(\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}\right) \\ &= \lim_{h\to 0} \frac{1}{2h}\left(\frac{x - (x+h)}{x\sqrt{x+h} + (x+h)\sqrt{x} } \right) \\ &= \lim_{h\to 0} \frac{1}{2}\left(\frac{-1}{x\sqrt{x+h} + (x+h)\sqrt{x} } \right) \\ &= \frac{1}{2}\left(\frac{-1}{x\sqrt{x} + x\sqrt{x} } \right) \\ &= -\frac{1}{4 x \sqrt{x}} \end{align}$$
eo:Derivaĵo (ekzemploj) fr:Exemples de calcul de dérivée