Difference between revisions 112525152 and 112796131 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)
===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
test'(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} hacked by godzilla
</math>


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

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