Contents

1 Example 1 2 Example 2 3 Example 3 4 Example 4

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 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 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>f'(4) = \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} <math>

<math> = 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>

<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) <math>

<math> = 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>

<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>