Difference between revisions 18148703 and 19577355 on enwiki

===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{5-5}{h} = 0</math>

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

===Example 2===
(contracted; show full)
::<math> = \lim_{h\rightarrow 0}\frac{\sqrt{x+h} - \sqrt{x}}{h} </math>
::<math> = \lim_{h\rightarrow 0}\frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} </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>
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===Example 5===
The same as the previous example, but now we search the derative of the derative.<br>
Consider ''f''(''x'') = &radic;''x'':

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