Revision 102467107 of "Derivative (examples)" on enwiki

:''For more background on this topic, see [[derivative]].''

===Example 1===
Consider ''f''(''x'') = 5:

: <math>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</math>

The derivative of a [[constant function]] is [[0 (number)|zero]].

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

:<math>
\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}
</math>

The derivative and slope are equivalent.

The derivative is difficult to comprehend for its douchebag properties. You see if you can fully comprehend a derivative, any at all, you are a douchebag.

===Example 3===
Via differentiation, one can find the slope of a curve. Consider <math>f(x)=x^2</math>:

:{|
|-
|<math> f'(x)\, </math>
|<math>= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} </math>
|-
|
|<math> = \lim_{h\rightarrow 0}\frac{(x+h)^2 - x^2}{h} </math>
|-
|
|<math> = \lim_{h\rightarrow 0}\frac{x^2 + 2xh + h^2 - x^2}{h} </math>
|-
|
|<math> = \lim_{h\rightarrow 0}\frac{2xh + h^2}{h} </math>
|-
|
|<math> = \lim_{h\rightarrow 0}(2x + h) = 2x </math>
|}

For any point ''x'', the slope of the function <math>f(x)=x^2</math> is <math>f'(x)=2x</math>.

===Example 4===
Consider <math> f(x) = \sqrt{x} </math>:

:{|
|-
|<math> f'(x)\, </math>
|<math>= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} </math>
|-
|
|<math> = \lim_{h\rightarrow 0}\frac{\sqrt{x+h} - \sqrt{x}}{h} </math>
|-
|
|<math> = \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) </math>
|-
|
|<math> = \lim_{h\rightarrow 0}\frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})} </math>
|-
|
|<math> = \lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h} + \sqrt{x}} </math>
|-
|
|<math> = \frac{1}{2 \sqrt{x}} </math>
|}

===Example 5===
The same as the previous example, but now we search the derivative of the derivative.<br>
Consider <math> f(x) = \sqrt{x} </math>:

:<math>
\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{\left(\frac{1}{2 \sqrt{x+h}}-\frac{1}{2 \sqrt{x}}\right)(2 \sqrt{x+h}+2 \sqrt{x})}{h(2 \sqrt{x+h}+2 \sqrt{x})} \\
       &= \lim_{h\rightarrow 0} \frac{\frac{2 \sqrt{x}}{2 \sqrt{x+h}}-\frac{2 \sqrt{x+h}}{2 \sqrt{x}}}{h(2 \sqrt{x+h}+2 \sqrt{x})} \\
       &= -\frac{1}{4 x \sqrt{x}}
\end{align}
</math>


[[Category:calculus]] [[Category:Mathematical notation]]

[[eo:Derivaĵo (ekzemploj)]]
[[fr:Exemples de calcul de dérivée]]