MATH 145 – Calculus I
September 9, 2025
Recall that if we want to find the instantaneous rate of change at a point \(x = a\), we use the following definition:
For a function \(f\), the instantaneous rate of change of \(f\) at point \(x = a\) is the quantity
\[ IV_{t=a} = \lim\limits_{h \to 0} AV_{[a,a+h]} = \lim\limits_{h \to 0} \frac{f(a + h) - f(a)} {h}. \]
Using this definition, we can define the derivative of a function \(f\) with respect to \(x\) evaluated at \(x=a\):
Let \(f\) be a function and \(x = a\) a value in the function’s domain. We define the derivative of \(f\) with respect to \(x\) evaluated at \(x = a\), denoted \(f'(a)\), by the formula
\[ f'(a) = \lim\limits_{h \to 0} \frac{f(a + h) - f(a)} {h}. \]
provided this limit exists.
Let \(f(x) = 3x - 4\).
Let \(g(x) = x^2 + 2\).
Let \(h(x) = \dfrac{1}{x}\).
There are a few different interpretations of the derivative of \(f\) with respect to \(x\) at \(x = a\). (This is not an exhaustive list.)
Let \(g(x) = x^2 + 2\). \[ \begin{aligned} AV_{[1,5]}\, g(x) &= 6 \\ AV_{[1,3]}\, g(x) &= 4 \\ AV_{[1,2]}\, g(x) &= 3 \\ AV_{[1,1.5]}\, g(x) &= 2.5 \\ AV_{[1,1.25]}\, g(x) &= 2.25 \\ AV_{[1,1.1]}\, g(x) &= 2.1 \\ AV_{[1,1.01]}\, g(x) &= 2.01 \\ AV_{[1,1.001]}\, g(x) &= 2.001 \\ &\vdots \\ g'(1) &= 2 \end{aligned} \]
Let’s try that from the other direction. \[ \begin{aligned} AV_{[-3,1]}\, g(x) &= -2 \\ AV_{[-1,1]}\, g(x) &= 0 \\ AV_{[0,1]}\, g(x) &= 1 \\ AV_{[0.5,1]}\, g(x) &= 1.5 \\ AV_{[0.75,1]}\, g(x) &= 1.75 \\ AV_{[0.9,1}\, g(x) &= 1.9 \\ AV_{[0.99,1]}\, g(x) &= 1.99 \\ AV_{[0.999,1]}\, g(x) &= 1.999 \\ &\vdots \\ g'(1) &= 2 \end{aligned} \]
There is an applet that shows the way that the limit of the slope of the secant line approaches the slope of the tangent line.
What are some points on this graph where the derivative is negative?
What are some points on this graph where the derivative is positive?
What are some points on this graph where the derivative is zero?
A potato is placed in an oven, and the potato’s temperature \(F\) (in degrees Fahrenheit) at various points in time \(t\) is taken and recorded in the following table. Time \(t\) is measured in minutes.
| \(t\) | 0 | 15 | 30 | 45 | 60 | 75 | 90 |
|---|---|---|---|---|---|---|---|
| \(F(t)\) | 70 | 180.5 | 251 | 296 | 324.5 | 342.8 | 354.5 |
If \(F(t)\) is the temperature at time \(t\), estimate the instantaneous rate of change of the temperature at \(t = 45\).
The central difference approximation to the value of the derivative at \(x = a\) is given by
\[ f'(a) \approx \frac{f(a + h) - f(a - h)}{2h}. \]
This quantity measures the slope of the secant line to \(y = f(x)\) through the points \((a-h, f(a-h))\) and \((a+h, f(a+h))\).
A potato is placed in an oven, and the potato’s temperature \(F\) (in degrees Fahrenheit) at various points in time \(t\) is taken and recorded in the following table. Time \(t\) is measured in minutes.
| \(t\) | 0 | 15 | 30 | 45 | 60 | 75 | 90 |
|---|---|---|---|---|---|---|---|
| \(F(t)\) | 70 | 180.5 | 251 | 296 | 324.5 | 342.8 | 354.5 |
If \(F(t)\) is the temperature at time \(t\), estimate \(F'(45)\).
The temperature change \(T\) (in Farhenheit degrees), in a patient, that is generated by a dose \(q\) (in milliliters) of a drug is given by the function \(T = f(q)\).
\[ f'(a) = \lim\limits_{h \to 0} \frac{f(a + h) - f(a)} {(x + h) - x}. \]
The units of the derivative of a function \(f\) with respect to \(x\) evaluated at \(x=a\), where \(y = f(x)\), are \(\dfrac{\textrm{units of }y}{\textrm{units of }x}\).