Difference between revisions 85887604 and 85888871 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]].
(contracted; show full)|<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> f''(x)\,</math>
|<math>= \lim_{h\rightarrow 0}\frac{f'(x+h)-f'(x)}{h} </math>
|-
|
|<math> = \lim_{h\rightarrow<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}</math>
|-
|
|<math> = \lim_{h\rightarrow \\
       &= \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})}</math>
|-
|
|<math>  \\
       &= \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})}</math>
|-
|
|<math> = \lim_{h\rightarrow 0} \frac{\frac{x}{\sqrt{x} \sqrt{x+h}}-\frac{x+h}{\sqrt{x} \sqrt{x+h}}}{h(2 \sqrt{x+h}+2 \sqrt{x})}</math>
|-
|
|<math>  \\
       &= -\frac{-1}{4 x \sqrt{x}}⏎
\end{align}⏎
</math>⏎
|}

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

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