Difference between revisions 20306409 and 20306519 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) \, </math> |<math>= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} </math> ::|- | |<math> = \lim_{h\rightarrow 0}\frac{(x+h)^2 - x^2}{h} </math> ::|- | |<math> = \lim_{h\rightarrow 0}\frac{x^2 + 2xh + h^2 - x^2}{h} </math> ::|- | |<math> = \lim_{h\rightarrow 0}\frac{2xh + h^2}{h} </math> ::|- | |<math> = \lim_{h\rightarrow 0}(2x + h) = 2x </math>⏎ |} For any point ''x'', the slope of the function <math>f(x)=x^2</math> is <math>f'(x)=2x</math>. ===Example 4=== Consider ''f''(''x'') = √''x'': :<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]] All content in the above text box is licensed under the Creative Commons Attribution-ShareAlike license Version 4 and was originally sourced from https://en.wikipedia.org/w/index.php?diff=prev&oldid=20306519.
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