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Wize University Calculus 1 Textbook > Derivatives

Derivatives as a Rate of Change

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Derivatives as a Rate of Change

Derivative is a fancy word for rate of change. The derivative is of a function tells us how much the function is changing at any value we wish!

Instantaneous Rate of Change (Slope of the Tangent Line)

The derivative f′f'gives the instantaneous rate of change (geometrically the slope of the tangent line) of ff to the curve at any point where the function is differentiable.


Note: A function is differentiable on an interval only if it is continuous over that interval!

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Notation for Derivatives

The derivative of the function f(x)f(x)is denoted:
f′(x)=y′=dfdx=dydx=ddxf(x)\boxed{\quad f'(x) = y' =\frac{df}{dx}=\frac{dy}{dx} =\frac{d}{dx}f\left(x\right)\quad }

If a function is constant, it has no rate of change. Thus the derivative of a constant, cc, is zero
ddx(c)=0\boxed{\dfrac{d}{dx}(c)=0}



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Average Rate of Change

The average rate of change of a function y=f(x)y=f(x) from x=ax=a to x=bx=b is defined as

Average Rate of Change=f(b)−f(a)b−a\boxed{\text{Average Rate of Change}=\frac{f(b)-f(a)}{b-a}}

Geometrically, this is the slope of the secant drawn to the graph of f(x)f(x) over the interval [a,b][a,b].



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Example: Average Rate of Change

Suppose that a company’s total cost (in dollars) to carry out a production of xx units of its product is given by C(x)=x2512+10x+100C(x)=\displaystyle\frac{x^{2}}{51^{2}}+10x+100. Find the average rate of change of the total cost for the first 5151 units produced.


Average rate of change=C(51)−C(0)51−0=C(51)−C(0)51=(512512+10(51)+100)−(0+0+100)51=1+10(51)51=51151\begin{array}{l} \displaystyle\text{Average rate of change}=\frac{C(51)-C(0)}{51-0}\\ \\ =\displaystyle\frac{C(51)-C(0)}{51}\\ \\ =\displaystyle\frac{\left(\frac{51^2}{51^2}+10(51)+100\right)-(0+0+100)}{51}\\ \\\displaystyle=\frac{1+10(51)}{51}\\ \\ \displaystyle =\frac{511}{51}\\ \\ \end{array}


f′(x)f'(x) plot for a differentiable function is depicted as below. Find interval(s) for which f(x)f\left(x\right) is increasing.


Extra Practice
Derivatives and Tangent Lines
A certain particle is traveling through space and time. We are able to measure the position of the particle at certain time values. Although we can't measure the particle continuously, we know that the particles position function is differentiable. The table gives values of time for the differentiable function s(t) for 0≤t≤40\le t\le4. s(t) represents the position of the particle at time t.
Practice: Derivatives as a Rate of Change
f′(x)f'(x) plot for a differentiable function is depicted as below. Find interval(s) for which f(x)f\left(x\right) is increasing.
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
The population of Vancouver in 2000 was 450,000. In 2020, the population was 740,000. What is the average annual rate of change of the population of Vancouver?
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Sherry bought a car in 2010 for $60,000\$60,000. 5050 years later, she sold the car as a collector car for $98,000\$98,000. What is the average annual rate of change of the value of the car? (Ignore units)
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
An investment, PP, in $, grows with interest over time tt (in years) is given by the table of values below.
t01234P10001305152817741936\begin{array}{|r|c|c|c|c|c|}\hline \textbf{t}&0&1&2&3&4\\\hline \textbf{P}&1000&1305&1528&1774&1936\\\hline \end{array}
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
The amount of bacteria, AA, in mg, of a certain species over time tt (in minutes) is given by the table of values below.
t01234A150183204247283\begin{array}{|r|c|c|c|c|c|}\hline \textbf{t}&0&1&2&3&4\\\hline \textbf{A}&150&183&204&247&283\\\hline \end{array}
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Determine the average rate of the change of the function g(t)=−3t2t+5g(t)=\dfrac{-3t}{2t+5} over the interval 0≤t≤50\leq{}t\leq{}5.
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Determine the average rate of the change of the function g(t)=t2t2−6g(t)=\dfrac{t^2}{t^2-6} over the interval 0.5≤t≤20.5\leq{}t\leq{}2.
Intervals of Increase and Decrease: Rate of Change
f′(x)f'(x) plot for a differentiable function is depicted as below. Find interval(s) for which f(x)f\left(x\right) is increasing.
The average rate of change of the function f(x)=2cos⁡(x)f(x)=2\cos(x) over the interval [π6,π2][ \frac{\pi}{6},\frac{\pi}{2}] is
Derivatives: Rate of Change
In recent years, exports from Canada to the United States are on the rise. Let E(t)E(t) represent the amount of exports to the United States from Canada (in Billions of dollars) at time tt. Let tt be the number of years after 2000. Interpret the following statements without using the words "derivative" or "instantaneous rate of change".
E(2)=5E(2)=5
E′(3)=1E'(3)=1
Derivatives and Tangent Lines
A certain particle is traveling through space and time. We are able to measure the position of the particle at certain time values. Although we can't measure the particle continuously, we know that the particles position function is differentiable. The table gives values of time for the differentiable function s(t) for 0≤t≤40\le t\le4. s(t) represents the position of the particle at time t.
The average rate of change of the function f(x)=3sin(x)−1f(x)=3sin(x)-1 over the interval [0,π6][0, \frac{\pi}{6}] is