Difference between revisions 81350941 and 81350997 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)|-
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|<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 
''f''(''x'') = √''x''<math> f(x) = \sqrt{x} </math>:

:{|
|-
|<math> f''(x)\,</math>
|<math>= \lim_{h\rightarrow 0}\frac{f'(x+h)-f'(x)}{h} </math>
|-
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|<math> = \lim_{h\rightarrow 0} \frac{\frac{1}{2 \sqrt{x+h}}-\frac{1}{2 \sqrt{x}}}{h}</math>
|-
|
|<math> = \lim_{h\rightarrow 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>
|-
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|<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>
|-
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|<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>
|-
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|<math> = \lim_{h\rightarrow 0} \frac{\frac{-h}{\sqrt{x} \sqrt{x+h}}}{h(2 \sqrt{x+h}+2 \sqrt{x})}</math>
|-
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|<math> = \lim_{h\rightarrow 0} \frac{-1}{\sqrt{x} \sqrt{x+h} (2 \sqrt{x+h}+2 \sqrt{x})}</math>
|-
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|<math> = \lim_{h\rightarrow 0} \frac{-1}{2 \sqrt{x} (x+h) + 2 x \sqrt{x+h}}</math>
|-
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|<math> = \frac{-1}{4 x \sqrt{x}}</math>
|}

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

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