Difference between revisions 20020877 and 20306409 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===
Consider the graph of <math>f(x)=2x-3</math>. If the reader has an understanding of [[algebra]] and the [[Cartesian coordinate system]], the reader should be able to independently determine that this [[line (mathematics)|line]] has a slope of 2 at every point. Using the above quotient (along with an understanding of the [[limit (mathematics)|limit]], [[secant]], and [[tangent]]) one can determine the slope at (4,5):

:{|
|-
|<math>f'(4)    \,  </math>
|<math>=  \lim_{h\rightarrow 0}\frac{f(4+h)-f(4)}{h} </math>
::|-
|
|<math> =  \lim_{h\rightarrow 0}\frac{2(4+h)-3-(2\cdot 4-3)}{h} </math>
::|-
|
|<math> =  \lim_{h\rightarrow 0}\frac{8+2h-3-8+3}{h} </math>
::|-
|
|<math> =  \lim_{h\rightarrow 0}\frac{2h}{h} = 2 </math>⏎
|}

The derivative and slope are equivalent.

===Example 3===
Via differentiation, one can find the slope of a curve. Consider <math>f(x)=x^2</math>:

:<math> f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} </math>
(contracted; show full)
::<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]]