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For more background on this topic, see derivative.

Contents 1 Example 1 2 Example 2 3 Example 3 4 Example 4 5 Example 5

[edit] 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 zero.

[edit] 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 has a slope of 2 at every point. Using the above quotient (along with an understanding of the limit, secant, and tangent) one can determine the slope at (4,5):

<math>

\begin{align} f'(4) &= \lim_{h\to 0}\frac{f(4+h)-f(4)}{h} \\

     &= \lim_{h\to 0}\frac{2(4+h)-3-(2\cdot 4-3)}{h} \\
     &= \lim_{h\to 0}\frac{8+2h-3-8+3}{h}            \\
     &= \lim_{h\to 0}\frac{2h}{h}                    \\
     &= 2

\end{align} </math>

The derivative and slope are equivalent.

[edit] 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>.

[edit] Example 4

Consider <math> f(x) = \sqrt{x} </math>:

<math> f'(x)\, </math>	<math>= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} </math>
	<math> = \lim_{h\rightarrow 0}\frac{\sqrt{x+h} - \sqrt{x}}{h} </math>
	<math> = \lim_{h\rightarrow 0}\left(\frac{\sqrt{x+h} - \sqrt{x}}{h}\right) \left(\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\right) </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>

[edit] Example 5

The same as the previous example, but now we search the derivative of the derivative.
Consider <math> f(x) = \sqrt{x} </math>:

<math>

\begin{align} f(x) &= \lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h} \\

      &= \lim_{h\to 0} \frac{\frac{1}{2 \sqrt{x+h}}-\frac{1}{2 \sqrt{x}}}{h} \\
      &= \lim_{h\to 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})} \\
      &= \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})} \\
      &= -\frac{1}{4 x \sqrt{x}}

\end{align} </math>eo:Derivaĵo (ekzemploj) fr:Exemples de calcul de dérivée