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Calculus for Team-Based Inquiry Learning: 2024 Early Edition

TBIL Institute Fellows

Section 2.2 Derivatives analytically (DF2)

Subsection 2.2.1 Activities

Observation 2.2.1.

Recall that \(f'(a)\text{,}\) the derivative of \(f(x)\) at \(x=a\text{,}\) was defined as the limit as \(h \to 0\) of the difference quotient on the interval \([a,a+h]\) as in Definition 2.1.9. If \(f'(a)\) exists, then say that \(f(x)\) is differentiable at \(a\text{.}\) If for some open interval \((a,b)\text{,}\) we have that \(f'(x)\) exists for every point \(x\) in \((a,b)\text{,}\) then we say that \(f(x)\) is differentiable on \((a,b)\text{.}\)

Activity 2.2.2.

For the function \(f(x)=x-x^2\) use the limit definition of the derivative at a point to compute \(f'(2)\text{.}\)
  1. \(\displaystyle f'(2)= \displaystyle \lim_{h\to 0} \frac{(2+h)-(2+h)^2-2+4}{h}=-3\)
  2. The limit \(f'(2)= \displaystyle \lim_{h\to 0} \frac{(2+h)-(2+h)^2-2}{h}\) simplifies algebraically to \(\displaystyle \lim_{h\to 0} \frac{-3h - h^2}{h}\) which does not exist, thus \(f'(2)\) is not defined.
  3. The limit \(f'(2)= \displaystyle \lim_{h\to 0} \frac{(2+h)-(2+h)^2-2}{h}\) simplifies algebraically to \(\displaystyle \lim_{h\to 0} \frac{h -h^2}{h}\) which does not exist, thus \(f'(2)\) is not defined.
  4. \(\displaystyle f'(2) = \displaystyle \lim_{h\to 0} \frac{(2+h)-(2^2+h^2)-2+4}{h}= 1\)

Activity 2.2.3.

Consider the function \(f(x)=3-2x\text{.}\) Which of the following best summarizes the average rates of changes of on \(f\) on the intervals \([1,4]\text{,}\) \([3,7]\text{,}\) and \([5, 5+h]\text{?}\)
  1. The average rate of change on the intervals \([1,4]\) and \([3,7]\) are equal to the slope of \(f(x)\text{,}\) but the average rate of change of \(f\) cannot be determined on \([5,5+h]\) without a specific value of \(h\text{.}\)
  2. The average rate of change on the intervals \([1,4]\text{,}\) \([3,7]\text{,}\) and \([5, 5+h]\) are all different values.
  3. The average rate of change on the intervals \([1,4]\text{,}\) \([3,7]\text{,}\) and \([5,5+h]\) are all equal to \(-2\text{.}\)

Activity 2.2.4.

Can you find \(f'(\pi)\) when \(f(x)=3-2x\) without doing any computations?
  1. No, because we cannot compute the value \(f(\pi)\text{.}\)
  2. No, because we cannot compute the average rate of change on the interval \([\pi, \pi+h]\text{.}\)
  3. Yes, \(f'(\pi)=3\) because the intercept of the tangent line at any point is equal to the constant intercept of \(f(x)\text{.}\)
  4. Yes, \(f'(\pi)=-2\) because the slope of the tangent line at any point is equal to the constant slope of \(f(x)\text{.}\)

Definition 2.2.5.

Let \(f(x)\) be function that is differentiable on an open interval \((a,b)\text{.}\) The derivative function of \(f(x)\text{,}\) denoted \(f'(x)\text{,}\) is given by the limit
\begin{equation*} f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}. \end{equation*}
At any particular input \(x=a\text{,}\) the derivative function outputs \(f'(a)\text{,}\) the value of derivative at the point \(x=a\text{.}\)

Remark 2.2.6.

To specify the indendent variable of our function, we say that \(f'(x)\) is the derivative of \(f(x)\) with respect to \(x\text{.}\) For the derivative function of \(y=f(x)\) we also use the notation:
\begin{equation*} f'(x)=y'(x)=\frac{dy}{dx}=\frac{df}{dx}. \end{equation*}
The last type of notation is known as differential (or Leibniz) notation for the derivative.

Remark 2.2.7.

Notice that our notation for the derivative function is based on the name that we assign to the function along with our choice of notation for indendent and dependent variables. For example, if we have a differentiable function \(y=v(t)\text{,}\) the derivative function of \(v(t)\) with respect to \(t\) can be written as \(v'(t)=y'(t)=\frac{dy}{dt}=\frac{dv}{dt}\text{.}\)

Activity 2.2.8.

In this activity you will consider \(f(x)=-x^2+4\) and compute its derivative function \(f'(x)\) using the limit definition of the derivative function Definition 2.2.5.
(a)
What expression do you get when you simplify the difference quotient
\begin{equation*} \frac{f(x+h)-f(x)}{h}= \frac{(-(x+h)^2+4)-(-x^2+4)}{h}? \end{equation*}
  1. \(\displaystyle \frac{x^2 + h^2 +4 -x^2 -4}{h} = \frac{h^2}{h}\)
  2. \(\displaystyle \frac{-x^2 - h^2 +4 +x^2 -4}{h} = \frac{-h^2}{h}\)
  3. \(\displaystyle \frac{-x^2 -2xh - h^2 +4 +x^2 -4}{h} = \frac{-2xh - h^2}{h}\)
  4. \(\displaystyle \frac{x^2 + 2xh + h^2 +4 - x^2 -4}{h} = \frac{2xh + h^2}{h}\)
(b)
After taking the limit as \(h \to 0\text{,}\) which of the following is your result for the derivative function \(f'(x)\text{?}\)
  1. \(\displaystyle f'(x)=x \)
  2. \(\displaystyle f'(x)=-x \)
  3. \(\displaystyle f'(x)=2x \)
  4. \(\displaystyle f'(x)=-2x \)

Activity 2.2.9.

Using the limit definition of the derivative, find \(f'(x)\) for \(f(x)=-x^2+2x-4\text{.}\) Which of the following is an accurate expression for \(f'(x)\text{?}\)
  1. \(\displaystyle f'(x)=2x +2\)
  2. \(\displaystyle f'(x)=-2x \)
  3. \(\displaystyle f'(x)=-2x+2 \)
  4. \(\displaystyle f'(x)=-2x -2 \)

Activity 2.2.10.

Using the limit definition of the derivative, you want to find \(f'(x)\) for \(f(x)=\frac{1}{x}\text{.}\) We will do this by first simplifying the difference quotient and then taking the limit as \(h\to 0\text{.}\)
(a)
What expression do you get when you simplify the difference quotient
\begin{equation*} \frac{f(x+h)-f(x)}{h}= \frac{\frac{1}{x+h}-\frac{1}{x}}{h}? \end{equation*}
  1. \(\displaystyle \frac{\frac{1}{x+h}}{h} = \frac{1}{(x+h)h}\)
  2. \(\displaystyle \frac{\frac{h}{x+h}}{h} = \frac{h}{h(x+h)}\)
  3. \(\displaystyle \frac{\frac{x-(x+h)}{(x+h)x}}{h} = \frac{-h}{h(x+h)x}\)
  4. \(\displaystyle \frac{\frac{x-(x+h)}{(x+h)x}}{h} = \frac{-h^2}{(x+h)x}\)
  5. \(\displaystyle \frac{\frac{h}{(x+h)x}}{h} = \frac{h}{h(x+h)x}\)
(b)
After taking the limit as \(h \to 0\text{,}\) which of the following is your result for the derivative function \(f'(x)\text{?}\)
  1. \(\displaystyle f'(x)=0 \)
  2. \(\displaystyle f'(x)=1/x \)
  3. \(\displaystyle f'(x)=-1/x \)
  4. \(\displaystyle f'(x)=1/x^2 \)
  5. \(\displaystyle f'(x)=-1/x^2 \)

Activity 2.2.11.

Find \(f'(x)\) using the limit definition of the derivative. Then evaluate at \(x=8\text{.}\)
\begin{equation*} f(x) = x^{2} - 5 \, x - 5 \end{equation*}

Definition 2.2.12.

Once we have computed the first derivative \(f'(x)\text{,}\) the second derivative of \(f(x)\) is the first derivative of \(f'(x)\) or
\begin{equation*} f''(x) = \lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h}. \end{equation*}

Activity 2.2.13.

Consider the function \(f(x)=-x^2+2x-4\text{.}\) Earlier you saw that \(f'(x)=-2x+2\text{.}\) What is the second derivative of \(f(x)\text{?}\)
  1. \(\displaystyle f''(x)=2 \)
  2. \(\displaystyle f''(x)=-2 \)
  3. \(\displaystyle f''(x)=2x \)
  4. \(\displaystyle f''(x)=-2x \)

Remark 2.2.14.

The first derivative encodes information about the rate of change of the original function. In particular,
  • If \(f'\gt 0 \text{,}\) then \(f\) is increasing;
  • If \(f'\lt 0 \text{,}\) then \(f\) is decreasing;
  • If \(f' = 0 \text{,}\) then \(f\) has a horizontal tangent line (and it might have a max or min or it might just be changing pace).
The second derivative is the derivative of the derivative. It encodes information about the rate of change of the rate of change of the original function. In particular,
  • If \(f'' \gt 0 \text{,}\) then \(f'\) is increasing;
  • If \(f'' \lt 0 \text{,}\) then \(f'\) is decreasing;
  • If \(f'' = 0 \text{,}\) then \(f'\) has a horizontal tangent line (and it might have a max or min or it might just be changing pace).

Activity 2.2.15.

Consider the function \(f(x)=-x^2+2x-4\text{.}\) Earlier you saw that \(f'(x)=-2x+2\) and \(f''(x)=-2\text{.}\) What does this tell you about the graph of \(f(x)\) for \(x \gt 1\text{?}\)
  1. The graph is increasing and concave up
  2. The graph is increasing and concave down
  3. The graph is decreasing and concave up
  4. The graph is decreasing and concave down

Observation 2.2.16.

We have two ways to compute analytically the derivative at a point. For example, to compute \(f'(1)\text{,}\) the derivative of \(f(x)\) at \(x=1\text{,}\) we have two methods
  1. We can directly compute \(f'(1)\) by finding the difference quotient on the interval \([1,1+h]\) and then taking the limit as \(h\to 0\text{.}\)
  2. We can first find the derivative function \(f'(x)\) by computing the difference quotient on the interval \([x,x+h]\text{,}\) then taking the limit as \(h\to 0\text{,}\) and finally evaluating the expression for \(f'(x)\) at the input \(x=1\text{.}\)
The latter approach is more convenient when you want to consider the value of the derivative function at multiple points!

Activity 2.2.17.

Consider the function \(f(x) = \frac{1}{x^2}\text{.}\) You will find \(f'(1)\) in two ways!
(a)
Using the limit definition of the derivative at a point, compute the difference quotient on the interval \([1,1+h]\) and then take the limit as \(h\to 0\text{.}\) What do you get?
  1. -1
  2. 1
  3. 2
  4. -2
(b)
Now, using the limit definition of the derivative function, find \(f'(x)\text{.}\) Which of the following is your result for the derivative function \(f'(x)\text{?}\)
  1. \(\displaystyle f'(x)=-1/x^3 \)
  2. \(\displaystyle f'(x)=1/x^3 \)
  3. \(\displaystyle f'(x)=-2/x^3 \)
  4. \(\displaystyle f'(x)=2/x^3 \)
(c)
Make sure that your answers match! So if you plug in \(x=1\) in \(f'(x)\text{,}\) you should get the same number you got when you computed \(f'(1)\text{.}\)

Activity 2.2.18.

In this activity you will study (again!) the velocity of a ball falling under gravity. A ball is tossed vertically in the air from a window. The height of the ball (in feet) is given by the formula \(f(t) = 64-16(t-1)^2\text{,}\) where \(t\) is the seconds after the ball is launched. Recall that in Activity 2.1.1, you used numerical methods to approxmiate the instantaneous velocity of \(f(t)\) to calculate \(v(2)\text{!}\)
(a)
Using the limit definition of the derivative function, find the velocity function \(v(t)=f'(t)\text{.}\)
(b)
Using the velocity function \(v(t)\text{,}\) what is \(v'(1)\text{,}\) the instantaneous velocity at \(t=1\text{?}\)
  1. -32 feet per second
  2. 32 feet per second
  3. 0 feet per second
  4. -16 feet per second
  5. 16 feet per second
(c)
What behavior would explain your finding?
  1. After 1 second the ball is falling at a speed of 32 meters per second.
  2. After 1 second the ball is moving upwards at a speed of 32 meters per second.
  3. After 1 second the ball reaches its highest point and it stops for an instant.
  4. After 1 second the ball is falling at a speed of 16 meters per second.
  5. After 1 second the ball is moving upwards at a speed of 16 meters per second.

Observation 2.2.19.

A function can only be differentiable at \(x=a\) if it is also continuous at \(x=a\text{.}\) But not all continuous functions are differentiable: when we have a corner in the graph of a the function, the function is continuous at the corner point, but it is not differentiable at that point!

Activity 2.2.20.

In Observation 2.1.17, we said that a function is not differentiable when the limit that defines it does not exist. In this activity we will study differentiability analytically.
(a)
Consider the following continuous function
\begin{equation*} g(x) = \begin{cases} x + 2 \amp x \leq 2 \\ x^2 \amp x \gt 2 \end{cases} \end{equation*}
Consider the interval \([2,2+h]\text{.}\) When \(h \lt 0 \text{,}\) the interval falls under the first definition of \(g(x)\) and the derivative is always equal to 1. What is the derivative function for \(x\) values greater than 2? Show that at \(x=2\) the value of this derivative is not equal to 1 and so \(g(x) \) is not differentiable at \(x=2\text{.}\)
(b)
Consider the following discontinuous function
\begin{equation*} g(x) = \begin{cases} x + 2 \amp x \leq 2 \\ x \amp x \gt 2 \end{cases} \end{equation*}
On both sides of \(x=2\) it seems that the slope is the same, but this function is still not differentiable at \(x=2\text{.}\) Notice that \(g(2)=4\text{.}\) When \(h \gt 0 \text{,}\) the interval \([2,2+h]\)falls under the second definition of \(g(x)\text{,}\) but \(g(2)\) is always fixed at 4. Compute the difference quotient \(\frac{g(2+h)-g(2)}{h}\) assuming that \(h \gt 0 \) and notice that this does not simplify as expected! Moreover, if you take the limit as \(h\to 0\text{,}\) you will get infinity and not the expected slope of 1!
(c)
Consider the following function
\begin{equation*} g(x) = \begin{cases} ax + 2 \amp x \leq 2 \\ bx^2 \amp x \gt 2 \end{cases} \end{equation*}
where \(a,b\) are some nonzero parameters you will find. Find an equation in \(a,b\) that needs to be true if we want the function to be continuous at \(x=2\text{.}\) Also, find an equation in \(a,b\) that needs to be true if we want the function to be differentiable at \(x=2\text{.}\) Solve the system of two linear equations... you should find that \(a=-2\) and \(b=-1/2\) are the only values that make the function differentiable (and continuous!).

Subsection 2.2.2 Videos

Figure 44. Video for DF2
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