Difference between revisions 101171372 and 101171579 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)       &= \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>


===Monomial Proof===
{|
|-
|
| Given <math> f(x) = x^n\, </math>
|-
|
| <math> f'(x)\, </math>
| <math>= \lim_{h\to 0}\frac{(x+h)^n - x^n}{h} </math>
|-
|
| <math> = \lim_{h\to 0}\frac{\sigma^n_i=0 - x^n}{h} </math>
|-
|
| <math> = \lim_{h\to 0}\frac{x^2 + 2xh + h^2 - x^2}{h} </math>
|-
|
| <math> = \lim_{h\to 0}\frac{2xh + h^2}{h} </math>
|-
|
| <math> = \lim_{h\to 0}(2x + h) = 2x. </math>
|}⏎
⏎

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

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