finite mathematics & applied calculus topic summary:

Introduction to the derivative: Part 3 of 3: Rates of change

Introduction to the derivative: Part 3 of 3: Rates of change

Average rate of change | $\displaystyle = \frac{\Delta f}{\Delta x} = \frac{f(b) - f(a)}{b - a}$. |

$=$ Slope of line through points $P$ and $Q$ (see the interactive graphic). |

- Average rate of change $\displaystyle = \frac{f(a+h) - f(a)}{h}$.

Use the handy little utilities below to compute the average rate of change of any function you enter.

$\displaystyle \frac{f(b) - f(a)}{b - a}$ | $\displaystyle = \frac{f(4) - f(2)}{4 - 2}$ |

$\displaystyle = \frac{17 - 1}{2} = 8$. |

Let $f(x) = x^2-x$. Then the average rate of change of $f$ over the interval $[3, 3+h]$ is

$\displaystyle \frac{f(a+h) - f(a)}{h}$ | $\displaystyle = \frac{f(3+h) - f(3)}{h}$ |

$\displaystyle = \frac{[(3+h)^2 - (3+h)] - [3^2 - 3]}{h}$ | |

$\displaystyle = \frac{9 + 6h + h^2 - 3 - h - 9 + 3}{h}$ | |

$\displaystyle = \frac{6h + h^2 - h}{h} = \frac{h(6 + h - 1)}{h}$ | |

$\displaystyle = 6 + h - 1$ | |

Instantaneous rate of change | $\displaystyle = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h}$. |

- $\displaystyle f'(a) = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h}$.

- $f'(a)$ = Instantaneous rate of change of $f$ at the point $a$. Thus,
$f'(x)$ = Instantaneous rate of change of $f$ at the point $x$.

Hence, the derivative $f'(x)$*is a function of*$x$, as its value can change with $x$. - As $f'(a)$ is a limit, it may or may not exist. That is, the quantities $\frac{f(a+h) - f(a)}{h}$ may or may not approach a fixed number as $h$ approaches zero. If everything works out fine and the limit exists, then we say that $f$ is
**differentiable**at $a$. Otherwise, we say that $f$ is**not differentiable**at $a$. - To talk about $f'(a)$, we require that $a$ be an
**interior point**of the domain of $f$; that is, we require that $f(x)$ be defined for all $x$ in some open interval about $x=a$. Thus, for instance, we do not talk about $f'(a)$ if $a$ is an endpoint of the domain of $f$.

- $f'(2) = 4$.

If, say, $f(x)$ represents the annual profit of your company (in millions of dollars) and $x$ represents the year since January 2015, then the units of measurement of the instantaneous rate of change are

- A table of values
- A "quick approximation"

- $\dfrac{f(x+h) - f(x)}{h}$

- $\dfrac{f(x+0.0001) - f(x)}{0.0001} \qquad$ Forward difference quotient

- $\dfrac{f(x+0.0001) - f(x-0.0001)}{0.0002} \qquad$ Balanced difference quotient

Enter a function, enter the point $a$, adjust $h$ if you want, and press "Compute".

The difference quotient (with $a = 2$) is

- $\dfrac{\color{blue}{f(2+h)} - \color{indianred}{f(2)}}{h} = \dfrac{\color{blue}{2(2+h)^2-4(2+h)+1} - \color{indianred}{2(2)^2-4(2)+1}}{h}$

h | 1 | 0.1 | 0.01 | 0.001 |

Difference Quotient | 6 | 4.2 | 4.02 | 4.002 |

- $f'(2) \approx 4$.

$\dfrac{\color{blue}{f(2+0.0001)} - \color{indianred}{f(2)}}{0.0001} = \dfrac{\color{blue}{f(2.0001)} - \color{indianred}{f(2)}}{0.0001}$ |

$\qquad = \dfrac{\color{blue}{2(2.0001)^2-4(2.0001)+1} - \color{indianred}{2(2)^2-4(2)+1}}{h}$ |

$\qquad = 4.0002.$ |

$\dfrac{\color{blue}{f(2+0.0001)} - \color{indianred}{f(2-0.0001)}}{0.0002} = \dfrac{\color{blue}{f(2.0001)} - \color{indianred}{f(1.9999)}}{0.0002}$ |

$\qquad = \dfrac{\color{blue}{2(2.0001)^2-4(2.0001)+1} - \color{indianred}{2(1.9999)^2-4(1.9999)+1}}{h}$ |

$\qquad = 4.$ |

- $\displaystyle v_{\text{average}} = \frac{s(t+h) - s(t)}{h}$.

- $\displaystyle v(t) = s'(t) = \lim_{h \to 0}\frac{s(t+h) - s(t)}{h}$.

- Slope of secant line $\displaystyle = m_{\text{sec}} = \frac{f(a+h) - f(a)}{h}$

Slope of tangent line $\displaystyle = m_{\text{tan}} = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h} = f'(a)$

Slope of secant line through points where $x = 2$ and $x = 3$ | ||

= Average rate of change of $f$ over $[2, 3]$
= $6$ (Check on the (blue) average rate of change calculator above.) | ||

Slope of tangent line through the point where $x = 2$ | ||

= Instantaneous rate of change of $f$ at $x = 2$
= $4$ (Check on the (blue) average rate of change calculator above.) |

- Write down the definition of the derivative,
- $\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$.

- Substitute for $f(x+h)$ and $f(x)$
- You can use an actual value for $x$ if you are asked, say, to compute $f'(3)$, or just leave it as $x$ if you are asked for the derivative function $f'(x)$.

- Simplify the numerator in order to factor out an $h$. Then cancel the $h$s and take the limit to obtain the answer.
- Sometimes, you need to do more than just simplify the numerator...

$\displaystyle f'(x)$ | $\displaystyle {}= \lim_{h \to 0}\frac{\color{red}{f(x+h)} - \color{blue}{f(x)}}{h}$ |

$\displaystyle {}= \lim_{h \to 0}\frac{\color{red}{2(x+h)^2-4(x+h)+1} - \color{blue}{(2x^2-4x+1)}}{h}$ | |

$\displaystyle {}= \lim_{h \to 0}\frac{2x^2+4xh+2h^2-4x-4h+1-2x^2+4x-1}{h}$ | |

$\displaystyle {}= \lim_{h \to 0}\frac{4xh+2h^2-4h}{h}$ | |

$\displaystyle {}= \lim_{h \to 0}\frac{h(4x+2h-4)}{h}$ | |

$\displaystyle {}= \lim_{h \to 0}(4x+2h-4)$ | |

$\displaystyle {}= 4x-4$ |