Difference between revisions 20306686 and 20307058 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> = \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'') = &radic;''x'':

:
{|
|-
|<math> f''(x)  \,</math>
|<math>= \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{\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> = \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> = \frac{-1}{4 x \sqrt{x}}</math>
::|-
|
|<math> = \frac{1}{4 x \sqrt{x}}</math> ⏎
|}⏎
⏎
(<math>\sqrt{x}</math> has 2 answers that only differ in sign, so it doesn't matter which sign we put in front of the end result.)

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