Section4.3Derivative Rules

Using the definition of the derivative of a function is quite tedious. In this section we introduce a number of different shortcuts that can be used to compute the derivative. Recall that the definition of derivative is:

Given any number $x$ for which the limit

\begin{equation*} f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \end{equation*}

exists, we assign to $x$ the number $f'(x)\text{.}$

Next, we give some basic Derivative Rules for finding derivatives without having to use the limit definition directly.

Let $f(x)=c$ be a constant function. By the definition of derivative:

\begin{equation*} f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{c-c}{h}=\lim_{h\to 0}0=0\text{.} \end{equation*}

We can also see the above theorem from a geometric point of view. Recall that the graph of a constant function is a horizontal straight line in the standard Cartesian coordinate system (see Figure 4.4). Now the tangent line to a horizontal straight line at any point on this line coincides with the line itself. But a horizontal line has slope zero, and therefore the slope of the tangent line must be zero as well.

Example4.25. Derivative of a Constant Function.

The derivative of $f(x)=17$ is $f'(x)=0$ since the derivative of a constant is $0\text{.}$

We use the formula:

\begin{equation*} x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1}) \end{equation*}

which can be verified by multiplying out the right side. Let $f(x)=x^n$ be a power function for some positive integer $n\text{.}$ Then at any number $a$ we have:

\begin{equation*} \begin{split} f'(a)\amp=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\\ \amp=\lim_{x\to a}\frac{x^n-a^n}{x-a}\\ \amp=\lim_{x\to a}(x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1})\\ \amp=na^{n-1}\end{split}\text{.} \end{equation*}

It turns out that the Power Rule holds for any real number $n\text{;}$ however, the proof of the Power Rule for the general case is a bit more difficult to prove and will be omitted.

Example4.28. Derivative of a Power Function.

By the Power Rule, the derivative of $g(x)=x^4$ is $g'(x)=4x^3\text{.}$

For convenience let $g(x)=cf(x)\text{.}$ Then:

\begin{equation*} \begin{split} g'(x)\amp =\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}=\lim_{h\to 0}\frac{cf(x+h)-cf(x)}{h}\\ \amp =\lim_{h\to 0}c\left[\frac{f(x+h)-f(x)}{h}\right]=c\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=cf'(x), \end{split} \end{equation*}

where $c$ can be moved in front of the limit by the Limit Rules.

Example4.30. Derivative of a Multiple of a Function.

Find the derivative of

\begin{equation*} f(x) = 5x^{3}\text{.} \end{equation*}
Solution

Using the Constant Multiple Rule, we obtain

\begin{equation*} f'(x)= \frac{d}{dx} \left(5x^{3}\right) = 5\frac{d}{dx}\left(x^{3}\right)\text{.} \end{equation*}

From the Power Rule, we have

\begin{equation*} \frac{d}{dx}\left(x^{3}\right) = 3x^{2}\text{.} \end{equation*}

Putting together, we get that

\begin{equation*} \frac{d}{dx} \left(5x^{3}\right) = 5\left(3x^{2}\right) = 15x^{2}\text{.} \end{equation*}
Example4.31. Derivative of a Multiple of a Function.

Find the derivative of

\begin{equation*} f(x) = \frac{3}{\sqrt{x}}\text{.} \end{equation*}
Solution

Rewrite $\frac{1}{\sqrt{x}}$ as $x^{-1/2}\text{.}$ Now, we can again use the Constant Multiple and Power Rules:

\begin{equation*} \begin{split} f'(x) \amp = \frac{d}{dx} \left(3x^{-1/2}\right) \\ \amp = 3 \frac{d}{dx} \left(x^{-1/2}\right) \\ \amp = 3\left(-\tfrac{1}{2}x^{-3/2}\right) \\ \amp = -\frac{3}{2x^{3/2}} \end{split} \end{equation*}

We demonstrate the proof for the Sum Rule and leave the proof for the Difference Rule to the reader.

For convenience, let $r(x)=f(x)+g(x)\text{.}$ Then by the Limit Rules:

\begin{equation*} \begin{split} r'(x)=\amp \lim_{h\to 0}\frac{r(x+h)-r(x)}{h}\\[1ex] =\amp \lim_{h\to 0}\frac{\left[f(x+h)+ g(x+h)\right]-\left[f(x)+ g(x)\right]}{h}\\[1ex] =\amp \lim_{h\to 0}\left[\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}\right]\\[1ex] =\amp \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}+\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\[1ex] =\amp f'(x)+g'(x) \end{split} \end{equation*}
Example4.33. Derivative of a Sum and Difference of Functions.

Find the derivative of

\begin{equation*} f(x) = 4x^{5} + 3x^{4} - 8x^{2} + x + 3\text{.} \end{equation*}
Solution
\begin{equation*} \begin{split} f'(x) \amp = \frac{d}{dx} \left(4x^{5} + 3x^{4} - 8x^{2} + x + 3\right) \\ \amp = \frac{d}{dx} (4x^{5}) + \frac{d}{dx} (3x^{4}) - \frac{d}{dx} (8x^{2}) + \frac{d}{dx} (x) + \frac{d}{dx}(3) \\ \amp = 4\frac{d}{dx} (x^{5}) + 3\frac{d}{dx} (x^{4}) - 8\frac{d}{dx} (x^{2}) + \frac{d}{dx} (x) + \frac{d}{dx}(3) \\ \amp = 20x^{4} + 12x^{3} - 16x + 1 \end{split} \end{equation*}
Example4.34. Tangent Line.

Find the slope and an equation of the tangent line to the graph of $f(x) = 2x + \frac{1}{\sqrt{x}}$ at the point $(1,3)\text{.}$

Solution

The slope of the tangent line at any point on the graph of $f$ is given by

\begin{equation*} \begin{split} f'(x) \amp = \frac{d}{dx} \left(2x + \dfrac{1}{\sqrt{x}}\right) \\ \amp = \frac{d}{dx} \left(2x+x^{-1/2}\right) \\ \amp = \frac{d}{dx} (2x) + \frac{d}{dx} (x^{-1/2}) \\ \amp = 2 - \tfrac{1}{2}x^{-3/2} \\ \amp = 2 - \frac{1}{2x^{3/2}} \end{split} \end{equation*}

In particular, the slope of the tangent line to the graph of $f$ at $(1,3)$ (where $x=1$) is

\begin{equation*} f'(1)=2-\frac{1}{2(1)^{3/2}} = 2-\frac{1}{2} = \frac{3}{2}\text{.} \end{equation*}

Using the point-slope form of the equation of a line with slope $m = \tfrac{3}{2}$ and the point $(x_{1},y_{1}) = (1,3)\text{,}$ we see that an equation of the tangent line is

\begin{equation*} \begin{split} y-y_{1} \amp = m(x-x_{1}) \\ y - 3 \amp = \tfrac{3}{2}(x-1), \end{split} \end{equation*}

or, upon simplification,

\begin{equation*} y = \frac{3}{2}x + \frac{3}{2}\text{.} \end{equation*}
Example4.35. Rate of Change Application.

The demand function for a certain product is given by

\begin{equation*} p(x)=\frac{\sqrt{x}}{2} - \frac{x}{40} + 2000\text{,} \end{equation*}

where $p$ is the price measured in dollars and the quantity $x$ is measured in units.

1. Find the rate of change of price $p$ per thousand products with respect to quantity $x\text{.}$

2. How fast is the price changing with respect to $x$ when $x=25$ and $x=400\text{?}$ Interpret your result.

Solution
1. The rate of change of the price with respect to quantity is given by

\begin{equation*} \begin{split} p'(x)\amp = \frac{d}{dx} \left(\frac{\sqrt{x}}{2} - \frac{x}{40} + 2000\right)\\[1ex] \amp =\frac{d}{dx} \left(\frac{1}{2}x^{1/2} - \frac{1}{40} x + 2000\right)\\[1ex] \amp = \frac{1}{2}\frac{d}{dx}(x^{1/2}) - \frac{1}{40}\frac{d}{dx}(x) \\[1ex] \amp =\frac{1}{2}\left(\frac{1}{2}x^{-1/2}\right) - \frac{1}{40}\\[1ex] \amp = \frac{1}{4} x^{-1/2} - \frac{1}{40}. \end{split} \end{equation*}
2. When $x=25\text{,}$ we have

\begin{equation*} p'(25)=\frac{1}{4}(25)^{-1/2} - \frac{1}{40} = \frac{1}{20}-\frac{1}{40} = \frac{1}{40} = 0.025\text{.} \end{equation*}

This means that when $25$ products are demanded, one additional product demanded by consumers increases the price by $$0.025\text{.}$ When $x=400\text{,}$ we have \begin{equation*} p'(400) = \frac{1}{4}(400)^{-1/2} - \frac{1}{40} = \frac{1}{80}-\frac{1}{40} = -\frac{1}{80} = -0.0125\text{.} \end{equation*} This means that when $400$ products are demanded, one additional product demanded by consumers decreases the price by$$0.0125\text{.}$ The graph of $p$ is shown below. Notice that although the price is decreasing for large numbers of products demanded, the decrease is minimal.

Example4.36. Second Derivative.

Find the second derivative of $f(x)=5x^3+3x^2\text{.}$

Solution

We must differentiate $f(x)$ twice:

\begin{equation*} f'(x)=15x^2+6x\text{,} \end{equation*}
\begin{equation*} f''(x)=30x+6\text{.} \end{equation*}

For convenience let $r(x)=f(x)\cdot g(x)\text{.}$ Then:

\begin{equation*} \begin{array}{ccl} r'(x)\amp =\amp \ds{\lim_{h\to 0}\frac{r(x+h)-r(x)}{h}}\\ \\ ~\amp =\amp \ds{\lim_{h\to 0}\frac{f(x+h)g(x+h) - f(x)g(x)}{h}} \end{array} \end{equation*}

As in the previous proof, we want to separate the functions $f$ and $g\text{.}$ The trick is to add and subtract $f(x+h)g(x)$ in the numerator. Then by Limit Rules:

\begin{equation*} \begin{array}{ccl} r'(x)\amp =\amp \ds{\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}}\\ \\ ~\amp =\amp \ds{\lim_{h\to 0}\left[f(x+h)\frac{g(x+h)-g(x)}{h}+g(x)\frac{f(x+h)-f(x)}{h}\right]}\\ \\ ~\amp =\amp \ds{\lim_{h\to 0}f(x+h)\cdot\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}+\lim_{h\to 0}g(x)\cdot\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}}\\ \\ ~\amp =\amp f(x)g'(x)+g(x)f'(x) \end{array} \end{equation*}

Note: we have just proven that the derivative of the product of two functions is not given by the product of the derivatives of the functions. In other words,

\begin{equation*} \frac{d}{dx}\left[f(x)g(x)\right] \color{red}\neq \frac{d}{dx}f(x)\frac{d}{dx}g(x)\text{.} \end{equation*}
Example4.38. Derivative of a Product of Functions.

Find the derivative of $\ds h(x)=(3x-1)(2x+3)\text{.}$

Solution

One way to do this question is to expand the expression. Alternatively, we use the Product Rule with $f(x)=3x-1$ and $g(x)=2x+3\text{.}$ Note that $f'(x)=3$ and $g'(x)=2\text{,}$ so,

\begin{equation*} h'(x)=(3)\cdot(2x+3)+(3x-1)\cdot(2)=6x+9+6x-2=12x+7\text{.} \end{equation*}

The proof is similar to the previous proof but the trick is to add and subtract the term $f(x)g(x)$ in the numerator. We omit the details.

Example4.40. Derivative of a Quotient of Functions.

Find the derivative of $\ds h(x)=\frac{3x-1}{2x+3}\text{.}$

Solution

By the Quotient Rule (using $f(x)=3x-1$ and $g(x)=2x+3$) we have:

\begin{equation*} h'(x)=\frac{\frac{d}{dx}(3x-1)\cdot (2x+3)-(3x-1)\cdot\frac{d}{dx}(2x+3)}{(2x+3)^2} \end{equation*}
\begin{equation*} =\frac{3(2x+3)-(3x-1)(2)}{(2x+3)^2}=\frac{11}{(2x+3)^2}\text{.} \end{equation*}
Example4.41. Rate of Change Application.

The sales (in millions of dollars) of a DVD recording of a hit movie $t$ years from the date of release is given by

\begin{equation*} S(t) = \frac{5t}{t^{2}+1}\text{.} \end{equation*}
1. Find the rate at which the sales are changing at time $t\text{.}$

2. How fast are the sales changing at the time the DVDs are relased ($t=0$)? Two years from the date of release?

Solution
1. The rate at which the sales are changing at time $t$ is given by $S'(t)\text{.}$ Using the Quotient Rule, we obtain

\begin{equation*} \begin{split} S'(t) \amp = \frac{d}{dt}\left(\frac{5t}{t^{2}+1}\right) = 5\frac{d}{dt}\left(\frac{t}{t^{2}+1}\right) \\[1ex] \amp = 5\left(\frac{(t^{2}+1)(1) - t(2t)}{(t^{2}+1)^{2}}\right) \\[1ex] \amp = 5\left(\frac{t^{2}+1-2t^{2}}{(t^{2}+1)^{2}}\right)\\[1ex] \amp = \frac{5(1-t^{2})}{(t^{2}+1)^{2}}. \end{split} \end{equation*}
2. The rate at which the sales are changing at the time the DVDs are released is given by

\begin{equation*} S'(0)= \frac{5(1-0)}{(0+1)^{2}}=5\text{.} \end{equation*}

That is, they are increasing at the rate of $$5$ million per year. Two years from the date of release, the sales are changing at the rate of \begin{equation*} S'(2)= \frac{5(1-4)}{(4+1)^{2}}=-0.6\text{.} \end{equation*} That is, they are decreasing at the rate of$$600,000$ per year.

The graph of the function $S$ is shown below where $S(t)$ is in millions of dollars and $t$ is in years. Notice that after a spectacular rise, sales of the DVD begin to taper off.

Exercises for Section 4.3.

Find the derivatives of the following functions.

1. $\ds f(x) = x^{100}$

$\ds 100x^{99}$
Solution
\begin{aligned}f'(x) = \diff{}{x} x^{100} = 100 x^{100-1} = 100x ^{99} \end{aligned}
2. $\ds f(t) = t^{-100}$

$\ds -100t^{-101}$
Solution
\begin{aligned}f'(t) = \diff{}{t} t^{-100} = (-100) t^{-100 -1} = -100t^{-101} \end{aligned}
3. $\ds g(x) = {1\over x^5}$

$\ds -5x^{-6}$
Solution
\begin{aligned}g'(x) = \diff{}{x} \frac{1}{x^5} = \diff{}{x} x^{-5} = -5 x^{-5-1} = -5 x^{-6} = - \frac{5}{x^6} \end{aligned}
4. $\ds f(x) = x^\pi$

$\ds \pi x^{\pi-1}$
Solution
\begin{aligned}f'(x) = \diff{}{x} x^{\pi} = \pi x^{\pi -1} \end{aligned}
5. $\ds h(x) = x^{3/4}$

$\ds (3/4)x^{-1/4}$
Solution
\begin{aligned}h'(x) = \diff{}{x} x^{3/4} = \frac{3}{4} x^{3/4 -1 } = \frac{3}{4} x^{-1/4} \end{aligned}
6. $\ds g(s) = s^{-9/7}$

$\ds -(9/7)s^{-16/7}$
Solution
\begin{aligned}g'(s) = \diff{}{s} s^{-9/7} = -\frac{9}{7} s^{-9/7 -1} = -\frac{9}{7} s^{-16/7} \end{aligned}
7. $\ds f(x) = 5x^3+12x^2-15$

$\ds 15x^2+24x$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} \left(5x^3+12x^2-15\right)\\ \amp = \diff{}{x} 5x^3 + \diff{}{x} 12x^2 - \diff{}{x} 15 \\ \amp = 5\diff{}{x} x^3 + 12 \diff{}{x} x^2 - 0 \\ \amp = 5 (3x^{3-1}) + 12 (2x^{2-1}) \\ \amp = 15x^2 + 24x \end{aligned}
8. $\ds f(s) = -4s^5 + 3s^2 - 5/s^2$

$\ds -20s^4+6s+10/s^3$
Solution
\begin{aligned}f'(s) \amp = \diff{}{s} \left(-4s^5+3s^2-\frac{5}{s^2}\right)\\ \amp = \diff{}{s} (-4s^5) + \diff{}{s} 3s^2 - \diff{}{s} \frac{5}{s^2}\\ \amp = -4 \diff{}{s} s^5 + 3 \diff{}{s} s^2 - 5 \diff{}{s} s^{-2} \\ \amp = -4 (5s^{5-1}) + 3 (2s^{2-1}) - 5(-2 s^{-2-1})\\ \amp = -20 s^4 + 6s + 10 s^{-3} \\ \amp = -20s^4 +6s + \frac{10}{s^3} \end{aligned}
9. $\ds f(x) = 5(-3x^2 + 5x + 1)$

$\ds -30x+25$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} 5\left(-3x^2+5x+1\right)\\ \amp = 5\diff{}{x} \left(-3x^2+5x+1\right) \\ \amp = 5 \left( -3 \diff{}{x} x^2 + 5\diff{}{x} x + \diff{}{x} 1 \right)\\ \amp = 5\left(-3(2x) + 5 (1) + 0\right)\\ \amp = 5\left(-6x + 5\right)\\ \amp = -30x+25 \end{aligned}
10. $\ds f(t) = (t+1)(t^2+2t-3)$

$\ds 3t^2+6t-1$
Solution
\begin{aligned}f'(t) \amp = \diff{}{x} (t+1)(t^2+2t-3) \\ \amp = (t^2+2t-3) \diff{}{t} (t+1) + (t+1) \diff{}{t} (t^2+2t-3)\\ \amp = (t^2+2t-3) \left(\diff{}{t} t + \diff{}{t} 1\right) + (t+1) \left( \diff{}{t} t^2 + \diff{}{t} 2t - \diff{}{t} 3 \right)\\ \amp = (t^2+2t-3) (1 + 0) + (t+1) (2t+2-0) \\ \amp = (t^2+2t-3) + (t+1)(2t+2) \end{aligned}
11. $\ds h(x) = (x+1)(x^2+2x-3)^{-1}$

$\ds -\frac{x^2+2x+5}{(x^2+2x-3)^2}$
Solution
\begin{aligned}h'(x) \amp = \diff{}{x} (x+1)(x^2+2x-3)^{-1} \\ \amp = \diff{}{x} \frac{x+1}{x^2+2x-3}\\ \amp = \frac{(x^2+2x-3) \diff{}{x} (x+1) - (x+1) \diff{}{x} (x^2+2x-3)}{(x^2+2x-3)^2}\\ \amp = \frac{(x^2+2x-3) (1+0) - (x+1) (2x+2-0)}{(x^2+2x-3)^2}\\ \amp = \frac{(x^2+2x-3) - (x+1)(2x+2)}{(x^2+2x-3)^2} \end{aligned}
12. $\ds g(x) = x^3(x^3-5x+10)$

$\ds 3x^2(x^3-5x+10)+x^3(3x^2-5)$
Solution
\begin{aligned}g'(x) \amp = \diff{}{x} x^3(x^3-5x+10) \\ \amp = (x^3-5x+10)\diff{}{x} x^3 + x^3 \diff{}{x} (x^3 - 5x + 10) \\ \amp = (x^3-5x+10) (3x^{3-1}) + x^3 \left(\diff{}{x} x^3 - 5\diff{}{x}x + \diff{}{x} 10\right)\\ \amp = (x^3-5x+10) (3x^2) + x^3 (3x^2-5+0) \\ \amp = 3x^3(x^3-5x+10) + x^3(3x^2-5) \end{aligned}
13. $\ds g(s) = (s^2+5s-3)(s^5)$

$\ds s^4(7s^2+30s-15)$
Solution
\begin{aligned}g'(s) \amp = \diff{}{s} (s^2+5s-3)(s^5)\\ \amp = s^5 \diff{}{s} (s^2+5s-3) + (s^2+5s-3) \diff{}{s} s^5 \\ \amp = s^5 \left(\diff{}{s} s^2 + 5\diff{}{s} s - \diff{}{s} 3 \right) + (s^2+5s-3) (5s^{5-1}) \\ \amp = s^5 \left(2s^{2-1} + 5(1)- 0 \right) + (s^2+5s-3) (5s^4) \\ \amp = s^5 \left(2s+5\right) + (s^2+5s-3)(5s^4) \end{aligned}
14. $\ds f(x) = (x^2+5x-3)(x^{-5})$

$\ds\frac{-3x^2-20x+15}{x^6}$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} (x^2+5x-3)(x^{-5})\\ \amp = \diff{}{x} \frac{x^2+5x-3}{x^{5}} \\ \amp = \frac{ x^5 \diff{}{x} (x^2+5x-3) - (x^2+5x-3)\diff{}{x} x^5}{(x^5)^2} \\ \amp = \frac{x^5 \left(\diff{}{x} x^2 + 5\diff{}{x} x - \diff{}{x} 3\right) - (x^2+5x-3) (5x^{5-1})}{x^{10}} \\ \amp = \frac{x^5 (2x^{2-1} + 5x^{1-1} - 0) - (x^2+5x-3)(5x^4)}{x^{10}} \\ \amp = \frac{x^5 (2x + 5) - (x^2+5x-3)(5x^4)}{x^{10}} \\ \amp = \frac{x(2x+5) - 5(x^2+5x-3)}{x^6} \end{aligned}
15. $\ds h(x) = (5x^3+12x^2-15)^{-1}$

$\ds -\frac{3x(5x+8)}{(5x^3+12x^2-15)^2}$
Solution
\begin{aligned}h'(x) \amp = \diff{}{x} (5x^3+12x^2-15)^{-1} \\ \amp = \diff{}{x} \frac{1}{5x^3 + 12x^2 - 15} \\ \amp = \frac{(5x^3+12x^2-15) \diff{}{x} (1) - (1)\diff{}{x}(5x^3+12x^2-15)}{(5x^3+12x^2-15)^2}\\ \amp = \frac{(5x^3+12x^2-15) (0) - \left(5\diff{}{x} x^3 + 12\diff{}{x} x^2 - \diff{}{x} 15\right)}{(5x^3+12x^2-15)^2}\\ \amp = \frac{-\left(5(3x^2) + 12(2x) - 0\right)}{(5x^3+12x^2-15)^2}\\ \amp = \frac{-\left(15x^2+24x\right)}{(5x^3+12x^2-15)^2} \end{aligned}

Find an equation for the tangent line to $\ds f(x) = x^3/4 - 1/x$ at $x=-2\text{.}$

$y=13x/4+5$
Solution

Using the power rule, we find the derivative of $f$ to be

\begin{equation*} f'(x) = \diff{}{x} \left(\frac{x^3}{4}-x^{-1}\right)=\frac{3x^2}{4} - (-x^{-2}) = \frac{3x^2}{4}+\frac{1}{x^2}\text{.} \end{equation*}

Therefore,

\begin{equation*} f'(-2) = \frac{3(-2)^2}{4} + \frac{1}{(-2)^2} = \frac{13}{4}\text{,} \end{equation*}

which gives the slope of the tangent line at $x=-2\text{.}$ We further know that the tangent line passes throught the point $(-2, f(-2)) = \left(-2,-\frac{3}{2}\right)\text{:}$

\begin{equation*} y+\frac{3}{2} = \frac{13}{4} \left(x+2\right)\text{,} \end{equation*}

or

\begin{equation*} y = \frac{13}{4}x + 5\text{.} \end{equation*}

Find an equation for the tangent line to $\ds f(x)= 3x^2 - \pi ^3$ at $x= 4\text{.}$

$\ds y=24x-48-\pi^3$
Solution

Let's first find the slope of the tangent line to $f(x)=3x^{2}-\pi^{3}$ at $x=4\text{.}$

\begin{equation*} f'(x) = \diff{ }{x}\left(3x^{2}-\pi^{3}\right) = 3\diff{ }{x} x^{2} - 0 = 6x\text{.} \end{equation*}

Thus, $f'(4) = 6(4) = 24\text{.}$

The tangent line will pass through the point $\left(4,f(4)\right) = \left(4,48-\pi^{3}\right)\text{.}$ We can now use the point-slope formula to find the equation of the tangent line to $f$ at $x=4\text{:}$

\begin{equation*} \frac{y-f(4)}{x-4} = f'(4) \implies y = 24(x-4)+48-\pi^{3}\text{.} \end{equation*}

Suppose the position of an object at time $t$ is given by $\ds f(t)=-49 t^2/10+5t+10\text{.}$ Find a function giving the speed of the object at time $t\text{.}$ The acceleration of an object is the rate at which its speed is changing, which means it is given by the derivative of the speed function. Find the acceleration of the object at time $t\text{.}$

$-49t/5+5\text{,}$ $-49/5$
Solution

The position of an object is given by $f(t)\text{.}$ Therefore, the speed $s(t)$ of the object is

\begin{equation*} \begin{split} s(t) \amp = f'(t) = \diff{}{t} \left(-\frac{49}{10} t^2 + 5t + 10\right)\\ \amp = -\frac{49}{10} \diff{}{t} t^2 + 5\diff{}{t} t + \diff{}{t} 10\\ \amp = -\frac{49}{10} (2t) + 5(1) + 0\\ \amp = -\frac{49}{5} t + 5 \end{split} \end{equation*}

Furthermore, the acceleration of the object is

\begin{equation*} \begin{split} s'(t) \amp = f''(t) = \diff{}{t} \left(-\frac{49}{5} t + 5\right) \\ \amp = -\frac{49}{5} \diff{}{t} (t) + \diff{}{t} (5) \\ \amp = -\frac{49}{5} (1) + 0 \\ \amp = -\frac{49}{5} \end{split} \end{equation*}

Let $\ds f(x) =x^3$ and $c= 3\text{.}$ Sketch the graphs of $f\text{,}$ $cf\text{,}$ $f'\text{,}$ and $(cf)'$ on the same diagram.

Solution

We are given $f(x)=x^3$ and $c=3\text{.}$ Therefore,

\begin{equation*} f'(x) = 3x^2, \text{ and } (cf)' = 3f'(x) = 9x^2\text{.} \end{equation*}

We plot $f\text{,}$ $cf\text{,}$ $f'$ and $(cf)'$ below.

The general polynomial $P$ of degree $n$ in the variable $x$ has the form $\ds P(x)= \sum _{k=0 } ^n a_k x^k = a_0 + a_1 x + \ldots + a_n x^n\text{.}$ What is the derivative (with respect to $x$) of $P\text{?}$

$\ds\sum_{k=1}^n ka_kx^{k-1}$
Solution

Let $P(x) = \sum_{k=0}^{n} a_k x^k$ be the general polynomial of degree $n\text{.}$ To find the derivative $P'(x)\text{,}$ we use first use the sum rule:

\begin{equation*} \begin{split} P'(x) \amp = \diff{}{x} \sum_{k=0}^{n} a_k x^k \\ \amp = \diff{}{x} \left(a_0 + a_1 x+ \dots + a_n x^n\right) \\ \amp = \diff{}{x} a_0 + \diff{}{x} a_1 x + \dots + \diff{}{x} a_n x^n \end{split} \end{equation*}

The $a_{k}$ are constants, and so by the constant multiple rule we have

\begin{equation*} \diff{}{x} a_0 + \diff{}{x} a_1 x + \dots + \diff{}{x} a_n x^n = a_0 \diff{}{x}(1) + a_1 \diff{}{x} (x) + a_2 \diff{}{x} (x^2) + \dots + a_n\diff{}{x} (x^n) \end{equation*}

We can now apply the power rule to each term. This gives us

\begin{equation*} \begin{split} P'(x) \amp = a_0 \diff{}{x}(1) + a_1 \diff{}{x} (x) + a_2 \diff{}{x} (x^2) + \dots + a_n\diff{}{x} (x^n)\\ \amp = 0 + a_1 (1) + a_2 (2x) + \dots + a_n (nx ^{n-1}) \\ \amp = 1\cdot a_1 + 2 \cdot a_2 x + \dots + n \cdot a_n x^{n-1} \\ \amp = \sum_{k=1}^{n} k a_k x^{k-1} \end{split} \end{equation*}

Find a cubic polynomial whose graph has horizontal tangents at $(-2 , 5)$ and $(2, 3)\text{.}$

$\ds x^3/16-3x/4+4$
Solution

We wish to find a cubic polynomial whose graph has horizontal tangents at $(-2,5)$ and $(2,3)\text{.}$ We let $f(x) = ax^3 + bx^2 + cx + d$ be a general cubic polynomial. Then

\begin{equation*} f'(x) = a(3x^2) + b(2x) + c(1) + 0 = 3ax^2 + 2bx + c\text{.} \end{equation*}

Our function $f$ has the desired horizontal tangents if both $f'(-2) = 0$ and $f'(2) = 0\text{.}$ Therefore, we require:

\begin{equation*} 12a - 4b + c= 0, \text{ and } 12a + 4b + c = 0\text{.} \end{equation*}

This leaves 3 unknowns but only 2 equations. However, we further require that $f(-2) =5$ and $f(2) = 3\text{.}$ This means that:

\begin{equation*} -8a +4b -2c + d = 5 \text{ and } 8a + 4b + 2c + d = 3\text{.} \end{equation*}

We now have 4 equations and 4 unknowns. So we solve the system of equations. First, we take:

\begin{equation*} \begin{array}{cc} \amp 12 a - 4b + c \amp = 0\\ -\amp 12 a + 4b + c \amp = 0\\ +\amp -8a + 4b -2c + d \amp = 5\\ +\amp 8a + 4b + 2c + d \amp = 3\\ \hline \amp 2d \amp = 8 \end{array} \end{equation*}

Therefore, we require $d = 4\text{.}$ This leaves us with three unknowns. Now, notice that

\begin{equation*} \begin{array}{cc} \amp 12 a - 4b + c \amp = 0\\ -\amp 12 a + 4b + c \amp = 0\\ \hline \amp -16b \amp = 0 \end{array} \end{equation*}

Hence, $b=0\text{.}$ Next, we have

\begin{equation*} \begin{array}{cc} \amp 2(12a +c) \amp =0\\ +\amp -8a -2c + 4 \amp = 5\\ \hline \amp 16a \amp = 1\end{array} \end{equation*}

And so $a = \dfrac{1}{16}\text{.}$ Finally we require

\begin{equation*} 12\frac{1}{16} = -c \implies c = -\frac{3}{4}\text{.} \end{equation*}

Altogether, we found that the cubic polynomial

\begin{equation*} f(x) = \frac{x^3}{16} - \frac{3x}{4} + 4 \end{equation*}

has the desired tangent lines.

Prove that $\ds{d\over dx}(cf(x))= cf'(x)$ using the definition of the derivative.

Solution

We use the definition of the derivative.

\begin{equation*} \begin{split} \diff{}{x} \left(cf(x)\right) \amp = \lim_{h\to 0}\frac{cf(x+h) - cf(x)}{h}\\ \amp = \lim_{h\to 0} \frac{c\left(f(x+h) - f(x)\right)}{h} \\ \amp = c \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ \amp = cf'(x) \end{split} \end{equation*}

Suppose that $f$ and $g$ are differentiable at $x\text{.}$ Show that $f-g$ is differentiable at $x$ using the two linearity properties from this section.

Solution

If both $f$ and $g$ are differentiable, then both

\begin{equation*} \diff{f}{x} \text{ and } \diff{g}{x} \end{equation*}

exist. Since

\begin{equation*} \diff{}{x} \left(f-g\right) = \diff{f}{x} - \diff{g}{x} \end{equation*}

by the difference rule, this means that $\diff{}{x}(f-g)$ must also be differentiable.

Use the Product Rule to compute the derivative of $\ds f(x)=(2x-3)^2\text{.}$ Sketch the function. Find an equation of the tangent line to the curve at $x=2\text{.}$ Sketch the tangent line at $x=2\text{.}$

$f'=4(2x-3)\text{,}$ $y=4x-7$
Solution

We rewrite

\begin{equation*} f(x) = (2x-3)^2 = (2x-3)(2x-3)\text{.} \end{equation*}

Therefore, by the product rule,

\begin{equation*} \begin{split} f'(x) \amp = (2x-3) \diff{}{x}(2x-3) + (2x-3) \diff{}{x} (2x-3)\\ \amp = 2 (2x-3) \diff{}{x} (2x-3) \\ \amp = 2 (2x-3) \left(2\diff{}{x} x - 3\diff{}{x} 1 \right) \\ \amp = 2(2x-3) (2 - 0) \\ \amp = 4(2x-3) \end{split} \end{equation*}

At $x=2\text{,}$ the slope of the tangent line is therefore

\begin{equation*} f'(2) = 4(4 - 3) = 4\text{.} \end{equation*}

Since the point of tangency is $(2,f(2)) = (2,1)\text{,}$ the equation of the tangent line is

\begin{equation*} y -1 = 4(x-2)\text{.} \end{equation*}

We sketch $f(x)$ and the tangent line at $x=2$ below.

Suppose that $f\text{,}$ $g\text{,}$ and $h$ are differentiable functions. Show that $(fgh)'(x) = f'(x) g(x)h(x) + f(x)g'(x) h(x) + f(x) g(x) h'(x)\text{.}$

Solution

We wish to show that

$$(fgh)'(x) = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)\text{.}\label{eq_products}\tag{4.3.1}$$

First, let $p(x)=(gh)(x)\text{.}$ Then, by the product rule, $p'(x) = (gh)'(x) = g'(x)h(x) + g(x)h'(x)\text{.}$

We also note that $(fgh)'(x) = (fp)'(x)\text{,}$ which we can again expand using the product rule:

$$(fp)'(x) = f'(x)p(x) + f(x)p'(x)\text{.}\label{eq_sec}\tag{4.3.2}$$

Finally, to show the desired equality (4.3.1), we simply need to substitute $p(x) = (gh)(x) = g(x)h(x)$ and our result for $p'(x)$ into (4.3.2). We obtain,

\begin{equation*} \begin{split} (fgh)'(x) \amp = f'(x)g(x)h(x) + f(x)(g'(x)h(x) + g(x)h'(x)) \\ \amp = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x), \end{split} \end{equation*}

which is exactly the identity (4.3.1).

Compute the derivative of $\ds {x^3\over x^3-5x+10}\text{.}$

$\ds {3x^2\over x^3-5x+10}-{x^3(3x^2-5)\over (x^3-5x+10)^2}$
Solution

We first apply the quotient rule:

\begin{equation*} \begin{split} \diff{}{x} \left(\frac{x^3}{x^3-5x+10}\right) \amp = \frac{(x^3-5x + 10)\diff{}{x} (x^3) + x^3 \diff{}{x} (x^3-5x+10)}{(x^3)^3}\\ \amp = \frac{(x^3-5x + 10)\diff{}{x} (x^3) + x^3 \diff{}{x} (x^3-5x+10)}{x^6} \end{split} \end{equation*}

Next, we apply the sum/difference rule and the power rule:

\begin{equation*} \begin{split} \diff{}{x} \left(\frac{x^3}{x^3-5x+10} \right) \amp = \frac{(x^3-5x + 10)\diff{}{x} (x^3) - x^3 \diff{}{x} (x^3-5x+10)}{(x^3-5x+10)^2}\\ \amp = \frac{(x^3-5x+10) (3x^2) - x^3 \left(\diff{}{x} (x^3) - 5 \diff{}{x}x + \diff{}{x}10\right)}{(x^3-5x+10)^2}\\ \amp = \frac{(x^3-5x+10) (3x^2) - x^3 (3x^2- 5(1) + (0))}{x^6}\\ \amp = \frac{3x^2(x^3-5x+10) - x^3 (3x^2-5)}{(x^3-5x+10)^2} \end{split} \end{equation*}

After simplifying, we find that

\begin{equation*} \diff{}{x} \left(\frac{x^3}{x^3-5x+10}\right) = \frac{3x^2}{x^3-5x+10} - \frac{x^3 (3x^2-5)}{(x^3-5x+10)^2}\text{.} \end{equation*}

Compute the derivative of $\ds {x^2+5x-3\over x^5-6x^3+3x^2-7x+1}\text{.}$

$\ds {2x+5\over x^5-6x^3+3x^2-7x+1}-{(x^2+5x-3)(5x^4-18x^2+6x-7)\over(x^5-6x^3+3x^2-7x+1)^2}$
Solution

Let

\begin{equation*} f(x) =\frac{x^2+5x-3}{x^5-6x^3+3x^2-7x+1}\text{.} \end{equation*}

Now use the quotient rule:

\begin{equation*} \begin{split} f'(x) \amp = \frac{(x^5-6x^3+3x^2-7x+1) \diff{}{x} \left(x^2+5x-3\right) + \left(x^2+5x-3\right) \diff{}{x} \left(x^5-6x^3+3x^2-7x+1\right)}{(x^5-6x^3+3x^2-7x+1)^2} \end{split} \end{equation*}

Now, using the sum/difference rule and constant multiple rule, we rewrite

\begin{equation*} \diff{}{x} \left(x^2+5x-3\right) = \diff{}{x} x^2 + 5 \diff{}{x} x - \diff{}{x} 3, \text{ and } \end{equation*}
\begin{equation*} \diff{}{x}\left(x^5-6x^3+3x^2-7x+1\right) = \diff{}{x} x^5 - 6 \diff{}{x} x^3 + 3 \diff{}{x} x^2 - 7 \diff{}{x} x + \diff{}{x} 1\text{.} \end{equation*}

Now using the power rule, we calculate:

\begin{equation*} \diff{}{x} \left(x^2+5x-3\right) = 2x + 5(1) - 0, \text{ and } \end{equation*}
\begin{equation*} \diff{}{x}\left(x^5-6x^3+3x^2-7x+1\right) =5x^4 - 6(3x^2) + 3(2x) - 7(1) + 0\text{.} \end{equation*}

Altogether, we find that

\begin{equation*} f'(x) = \frac{(x^5-6x^3+3x^2-7x+1) (2x+5) + \left(x^2+5x-3\right) (5x^4-18x^2+6x-7)}{(x^5-6x^3+3x^2-7x+1)^2}\text{.} \end{equation*}

Compute the derivative of $\ds {x\over(x-625)^2}\text{.}$

$\ds \frac{x-1250}{2(x-625)^{3/2}}$
Solution

We first apply the quotient rule:

\begin{equation*} \diff{}{x} \left(\frac{x}{(x-625)^2}\right) = \frac{(x-625)^2 \diff{}{x} x - x\diff{}{x} (x-625)^2}{\left((x-625)^2\right)^2} \end{equation*}

From the power rule, we know that

\begin{equation*} \diff{}{x} x = 1\text{,} \end{equation*}

but we cannot immediately use the power rule to deduce

\begin{equation*} \diff{}{x} (x-625)^2 \end{equation*}

Therefore, we rewrite $(x-625)^2=(x-625)(x-625)$ and apply the product rule:

\begin{equation*} \diff{}{x} (x-625)(x-625) = (x-625)\diff{}{x} (x-625) + (x-625) \diff{}{x} (x-625) = 2(x-625) \end{equation*}

Therefore, we have

\begin{equation*} \diff{}{x} \left(\frac{x}{(x-625)^2}\right) = \frac{(x-625)^2 - 2x(x-625)}{(x-625)^4}\text{.} \end{equation*}

Compute the derivative of $\ds {(x-5)^2\over x^{20}}\text{.}$

$\ds \frac{200-39x}{2x^{21}\sqrt{x-5}}$
Solution

Applying the quotient rule gives

\begin{equation*} \diff{}{x} \frac{(x-5)^2}{x^{20}} = \frac{x^{20}\diff{}{x} (x-5)^2 - (x-5)^2\diff{}{x} x^{20}}{(x^{20})^2}\text{.} \end{equation*}

Next, using the power rule, we have

\begin{equation*} \diff{}{x} x^{20} = 20 x^{19}\text{,} \end{equation*}

and using the product rule, we see that

\begin{equation*} \diff{}{x} (x-5)^2 = \diff{}{x} (x-5) (x-5) = (x-5)\diff{}{x} (x-5) + (x-5)\diff{}{x}(x-5) = 2(x-5)\text{.} \end{equation*}

Putting together and simplifying, we see that

\begin{equation*} \diff{}{x} \frac{(x-5)^2}{x^{20}} = \frac{2x^{20}(x-5) - 20(x-5)^2x^{19}}{x^{40}} = \frac{2x(x-5) - 20(x-5)^2}{x^{21}} = \frac{2(x-5)(9x-50)}{x^{21}}\text{.} \end{equation*}

Find an equation for the tangent line to $\ds f(x) = (x^2 - 4)/(5-x)$ at $x= 3\text{.}$

$\ds y=17x/4-41/4$
Solution

We first differentiate $f(x)\text{:}$

\begin{equation*} \begin{split} f'(x) \amp = \diff{}{x} \frac{x^2-4}{5-x}\\ \amp = \frac{(5-x)\diff{}{x} (x^2-4) - (x^2-4)\diff{}{x} (5-x)}{(5-x)^2}\\ \amp = \frac{(5-x) \left(\diff{}{x} x^2 - \diff{}{x} 4\right) - (x^2-4) \left(\diff{}{x} 5 - \diff{}{x} x\right)}{(5-x)^2}\\ \amp = \frac{(5-x)(2x-0) - (x^2-4)(0 - 1)}{(5-x)^2}\\ \amp = \frac{2x(5-x) + (x^2-4)}{(5-x)^2}\\ \amp = \frac{-(x^2-10x+4)}{(5-x)^2} \end{split} \end{equation*}

Therefore, at the point $(3,f(3)) = \left(3,\frac{5}{2}\right)\text{,}$ the slope of the tangent line is

\begin{equation*} f'(3) = \frac{- (9 - 30 + 4)}{2^2} = \frac{17}{4}\text{.} \end{equation*}

Hence, an equation of the tangent line is

\begin{equation*} y - \frac{5}{2} = \frac{17}{4} (x-3)\text{.} \end{equation*}

Find an equation for the tangent line to $\ds f(x) = (x-2)/(x^3 + 4x - 1)$ at $x=1\text{.}$

$y=11x/16-15/16$
Solution

We first differentiate:

\begin{equation*} \begin{split} f'(x) \amp = \diff{}{x} \frac{x-2}{x^3+4x-1}\\ \amp = \frac{(x^3+4x-1)\diff{}{x} (x-2) - (x-2) \diff{}{x} (x^3+4x-1)}{(x^3+4x-1)^2}\\ \amp = \frac{(x^3+4x-1) \left(\diff{}{x} x - \diff{}{x} 2\right) - (x-2) \left(\diff{}{x} x^3 + 4\diff{}{x} x - \diff{}{x} 1 \right)}{(x^3+4x-1)^2}\\ \amp = \frac{(x^3+4x-1) \left(1 - 0\right) - (x-2) \left(3x^2 + 4(1) - 0 \right)}{(x^3+4x-1)^2}\\ \amp = \frac{(x^3+4x-1)- (x-2) \left(3x^2 + 4\right)}{(x^3+4x-1)^2}\\ \amp = \frac{-2x^3+6x^2+7}{(x^3+4x-1)^2} \end{split} \end{equation*}

The point of tangency at $x=1$ is $(1, f(1)) = \left(1, \frac{-1}{4}\right)\text{,}$ and the slope of the tangent line is

\begin{equation*} f'(1) = \frac{-2 + 6+7}{(1+4-1)^2} = \frac{11}{16}\text{.} \end{equation*}

Therefore, an equation of the tangent line is

\begin{equation*} y + \frac{1}{4} = \frac{11}{16} \left(x-1\right)\text{.} \end{equation*}

If $f'(4) = 5\text{,}$ $g'(4) = 12\text{,}$ $(fg)(4)= f(4)g(4)=2\text{,}$ and $g(4) = 6\text{,}$ compute $f(4)$ and $\ds{d\over dx}{f\over g}$ at 4.

$13/18$
Solution

We are given that $f(4)\cdot g(4) = 2$ and $g(4) = 6\text{,}$ and so we conclude that $f(4)=\frac{1}{3}\text{.}$ We now use the quotient rule:

\begin{equation*} \begin{split} \diff{}{x} \left(\frac{f(x)}{g(x)}\right) \bigg\vert_{x=4} \amp = \frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2} \bigg\vert_{x=4} \\ \amp = \frac{f'(4)g(4)-g'(4)f(4)}{\left(g(4)\right)^2}\\ \amp = \frac{5\cdot 6 - 12\cdot \frac{1}{3}}{6^2} = \frac{13}{18} \end{split} \end{equation*}

Let $f(x)=x^{3}\text{.}$ Find the point on the graph of $f$ where the tangent line is horizontal. Sketch the graph of $f$ and draw the horizontal tangent line.

Solution

Since $f'(x) = 3x^2$ (by the power rule), we see that $f'(x) = 0 \iff x = 0\text{.}$ Therefore, the tangent line to $f$ is horizontal at the point $(0,0)\text{:}$

Let $f(x)=x^{3}+1\text{.}$

1. Find the point(s) on the graph of $f$ where the slope of the tangent line is equal to 12.

$(-2,-7)$ and $(2,9)\text{.}$
Solution

The slope of the tangent line an any point $(x,f(x))$ on the graph of $f=x^{3}+1$ is given by

\begin{equation*} f'(x) = 3x^{2}\text{.} \end{equation*}

At the point(s) where the slope is equal to $12\text{,}$ we have $3x^{2} = 12\text{,}$ or $x = \pm 2\text{.}$ So, we get two points: $(-2,f(-2)) = (-2,-7)$ and $(2,f(2)) = (2,9)\text{.}$

2. Find the equation(s) of the tangent line(s) of part (a).

$y=12x+17$ and $y=12x-15\text{.}$
Solution

The equation of the tangent line at $(-2,-7)$ is

\begin{equation*} \frac{y+7}{x+2} = 12 \implies y = 12x + 17\text{,} \end{equation*}

and the equation of the tangent line at $(2,9)$ is

\begin{equation*} \frac{y-9}{x-2} = 12 \implies y = 12x -15\text{.} \end{equation*}
3. Sketch the graph of $f$ showing the tangent line(s).

We sketch the graph of $f(x)$ and the two tangent lines found above:

Let $f(x)=\frac{2}{3}x^{3}+x^{2}-12x+6\text{.}$ Find the values of $x$ for which

1. $f'(x)=-12$

$x=0$ or $x=-1\text{.}$
Solution
\begin{aligned}-12 \amp = 2(x^2+x-6) \\ 0 \amp = x^2 + x \\ 0 \amp = x(x+1) \\ \implies x \amp =0,-1 \end{aligned}
2. $f'(x)=0$

$x=-3$ or $x=2\text{.}$
Solution
\begin{aligned}0 \amp = 2(x^2+x-6) \\ 0 \amp = (x+3)(x-2) \\ \implies x \amp = -3,2 \end{aligned}
3. $f'(x)=12$

$x=-4$ or $x=3\text{.}$
Solution
\begin{aligned}12 \amp = 2(x^2+x-6) \\ 0 \amp = x^2+x-12 \\ 0 \amp = (x+4)(x-3) \\ \implies x \amp = -4,3 \end{aligned}

An economy's consumer price index (CPI) is described by the function

\begin{equation*} I(t)=-0.2t^{3}+3t^{2}+100 \ \ \ 0 \leq t \leq 10 \end{equation*}

where $t=0$ corresponds to $1994\text{.}$

1. At what rate was the CPI changing in $1999\text{?}$ In $2001\text{?}$ In $2004\text{?}$

$12.6$ pts/yr; $0$ pts/yr.
Solution

We compute the derivative using the power rule:

\begin{equation*} I'(t) = \diff{}{t} \left(-0.2t^3+3t^2+100\right) = -0.2(3)t^2+3(2)t+ 0 = -0.6t^2+6t = 6t\left(1-\frac{t}{10}\right) \end{equation*}

Therefore, in 1999, $t=5$ and $I'(5) = \frac{30}{2} = 15\text{.}$ That is, the CPI was increasing at a rate of $15$ points per year. In 2001, $t=7$ and $I'(7) = 12.6\text{.}$ That is, the CPI was increasing at a rate of $12.6$ points per year. In 2004, $t=10$ and $I'(10) = 0$ points per year. That is, the CPI was stationary.

2. What was the average rate of increase in the CPI over the period from $1999$ to $2004\text{?}$

$10$ pts/yr
Solution

The average rate of increase from 1999 to 2004 was

\begin{equation*} \frac{I(10) - I(5)}{5} = \frac{200-150}{5} = 10\text{.} \end{equation*}

points per year.

The demand function for the Luminar desk lamp is given by

\begin{equation*} p=f(x)=-0.1x^{2}-0.4x+35 \end{equation*}

where $x$ is the quantity demanded (measured in thousands) and $p$ is the unit price in dollars.

1. Find $f'(x)\text{.}$

$f'(x)=-0.2x-0.4\text{.}$
Solution

We differentiate:

\begin{equation*} \begin{split} f'(x) \amp = \diff{}{x} \left(-0.1x^2-0.4x + 35 \right)\\ \amp = -0.1 \diff{}{x} x^2 - 0.4 \diff{}{x} x + \diff{}{x} 35 \\ \amp = -0.1 (2x) - 0.4 (1) + 0 \\ \amp = -\frac{1}{10} \left(2x + 4\right) \end{split} \end{equation*}
2. What is the rate of change of the unit price when the quantity demanded is $10,000$ units ($x=10$)? What is the unit price at that level of demand?

$$2.40$ per $1000$ lamps;$$21$
Solution

When $x=10\text{,}$ $p'(10) = -\dfrac{24}{10} = -2.4\text{.}$ The unit price when $x=10$ is set at $p(10) = -10-4+35 = 21\text{.}$ Therefore, when 10,000 lamps are demanded, the unit price is decreasing at a rate of $2.40 per 10,000 lamps from$21.

The supply function for a certain make of transistor radio is given by

\begin{equation*} p=f(x)=0.00001x^{5/4}+10 \end{equation*}

where $x$ is the quantity supplied and $p$ is the unit price in dollars.

1. Find $f'(x)\text{.}$

$f'(x)=0.000125x^{1/4}\text{.}$
Solution

We differentiate:

\begin{equation*} f'(x) = \diff{}{x} \left(0.00001x^{5/4} + 10\right) = 0.0000125 x^{1/4}\text{.} \end{equation*}
2. What is the rate of change of the unit price if the quantity supplied is $10,000$ radios?

$$2.08$million/yr. Solution When the unit price is 10,000 radios, $x=10,000\text{.}$ So \begin{equation*} f'(10,000) = 0.0000125(10,000)^{1/4} = 0.000125, \text{ and } \end{equation*} \begin{equation*} f(10,000) = 11\text{.} \end{equation*} That is, the unit price is increasing at a rate of approximately 0.000125 per 10,000 radios from a value of$11.

Despite efforts at cost containment, the cost of medical care is increasing. Two major reasons for this increase are an aging population and extensive use by physicians of new technologies. Health-care spending through the year $2000$ may be approximated by

\begin{equation*} S(t)=0.02836t^{3}-0.05167t^{2}+9.60881t+41.9 \ \ \ 0\leq t \leq 35 \end{equation*}

where $S(t)$ is the spending in millions of dollars and $t$ is measured in years, with $t=0$ corresponding to the beginning of $1965\text{.}$

1. Find an expression for the rate of change of health-care spending at any time $t\text{.}$

$S'(t)=0.08508t^{2}-0.10334t+9.60881$
Solution

The rate of change of health-care spending with respect to time is $S'(t)\text{:}$

\begin{equation*} S'(t) = 3(0.02836)t^2 - 2(0.05167)t + 9.60881 = 0.08508 t^2 - 0.10334t + 9.60881 \end{equation*}
2. How fast was health-care spending changing at the beginning of $1980\text{?}$ At the beginning of $2000\text{?}$

$$27.20171$ million/yr;$$110.21491$ million/yr
Solution

The year $1980$ corresponds to $t=15\text{,}$ therefore

\begin{equation*} S'(15) = 27.2017\text{.} \end{equation*}

That is, health-care spending was increasing at a rate of about 27.2 million dollars per year at the beginning of 1980. Similarly, at the beginning of the year 2000, $t=35$ and

\begin{equation*} S'(35) = 110.215\text{.} \end{equation*}

That is, health-care spending was increasing at a rate of about 110.2 million dollars per year.

3. What was the amount of health-care spending at the beginning of $1980\text{?}$ At the beginning of $2000\text{?}$

$$270.1214$ million;$$1530.8476$ million
Solution

Health-care spending at the beginning of 1980 was

\begin{equation*} S(15) = 270.121\text{,} \end{equation*}

or about 270.121 million dollars. At the beginning of 2000, health-care spending was

\begin{equation*} S(35) = 1530.85\text{,} \end{equation*}

Find the point(s) on the graph of the function

\begin{equation*} f(x)=(x^{2}+6)(x-5) \end{equation*}

where the slope of the tangent line is equal to $-2\text{.}$

$(\frac{4}{3},-\frac{770}{27})$ and $(2,-30)\text{.}$
Solution

We rewrite $f(x)=(x^2+6)(x-5)$ as $f(x)=x^3-5x^2+6x-30\text{.}$ Then we differentiate using the power rule:

\begin{equation*} f'(x) = \diff{}{x} \left(x^3-5x^2+6x-30\right) = 3x^2-10x+6\text{.} \end{equation*}

So, the slope of the tangent line is equal to -2 when

\begin{equation*} -2 = 3x^2-10x+6 \implies 0 = 3x^2-10x+8\text{.} \end{equation*}

Using the quadratic formula, we find

\begin{equation*} x = \frac{4}{3}, 2\text{.} \end{equation*}

Therefore, the slope of the tangent line to $f$ is equal to -2 at the point

\begin{equation*} \left(\frac{4}{3}, f\left(\frac{4}{3}\right)\right) = \left(\frac{4}{3},-\frac{770}{27}\right)\text{,} \end{equation*}

and at the point

\begin{equation*} \left(2,f(2)\right) = \left(2, -30\right)\text{.} \end{equation*}

A straight line perpendicular to and passing through the point of tangency of the tangent line is called the normal to the curve. Find the equation of the tangent line and the normal to the curve $y = \frac{1}{1+x^{2}}$ at the point $(1,\frac{1}{2})\text{.}$

$y=-\frac{1}{2}x+1$ and $y=2x-\frac{3}{2}\text{.}$
Solution

We first find the equation of the tangent line. We differentiate:

\begin{equation*} \begin{split} \diff{}{x} \frac{1}{1+x^2} \amp = \frac{(1+x^2)\diff{}{x}(1) - \diff{}{x} (1+x^2)}{(1+x^2)^2} \\ \amp = \frac{(1+x^2)(0) - \left(\diff{}{x} (1) + \diff{}{x} x^2\right)}{(1+x^2)^2}\\ \amp = \frac{-2x}{(1+x^2)^2} \end{split} \end{equation*}

So at the point $(1,1/2)\text{,}$ the slope of the tangent line is

\begin{equation*} \frac{-2(1)}{(1+(1)^2)^2} = \frac{-2}{2^2} = -\frac{1}{2}\text{.} \end{equation*}

And so an equation of the tangent line is

\begin{equation*} y - \frac{1}{2} = -\frac{1}{2} (x-1)\text{.} \end{equation*}

Now to find an equation of the normal line, we use the fact that it is perpendicular to the tangent line. Therefore, it has slope $2\text{.}$ Since this line also passes through the point $(1,1/2)\text{,}$ an equation of the normal line is

\begin{equation*} y - \frac{1}{2} = 2 (x-1)\text{.} \end{equation*}

The total revenue in dollars for a video game is given by

\begin{equation*} R(x)=\tfrac{1}{100}(x+2000)(1600-x)-36,000 \end{equation*}

where $x$ is the number of units sold. What is the rate of change of revenue with respect to $x$ when $600$ units are sold? Interpret your result.

$R'(600)=-16\text{.}$ The revenue is decreasing at a rate of $$16$ per video game. Solution We differentiate the total revenue with respect to $x\text{:}$ \begin{equation*} \begin{split} R'(x) \amp = \diff{}{x} \left(\frac{1}{100} (x+2000)(1600-x)-36,000\right)\\ \amp = \frac{1}{100} \diff{}{x} \left((x+2000)(1600-x)-36,000\right)\\ \amp = \frac{1}{100} \left(\diff{}{x} (x+2000)(1600-x) - \diff{}{x} 36,000\right) \\ \amp = \frac{1}{100} \left((x+2000)\diff{}{x}(1600-x) + (1600-x)\diff{}{x}(x+2000)\right) \\ \amp = \frac{1}{100} \left((x+2000)(-1) + (1600-x)(1)\right)\\ \amp = \frac{1}{100} \left(-(x+2000) + (1600-x)\right) \end{split} \end{equation*} Therefore, when $x=600\text{,}$ $R'(600) = -16\text{.}$ So the revenue is decreasing at a rate of$16 per unit sold.

A city's main well was recently found to be contaminated with trichloroethylene, a cancer-causing chemical, as a result of an abandoned chemical dump leaching chemicals into the water. A proposal submitted to the city's council members indicates that the cost, measured in millions of dollars, of removing $x$% of the toxic pollutant is given by

\begin{equation*} C(x) = \frac{0.5x}{100-x}\text{.} \end{equation*}

Find $C'(80), C'(90), C'(95), C'(99)\text{.}$ What do your results tell you about the cost of removing all of the pollutant?

$$0.125\text{,}$$$0.5\text{,}$$$2\text{,}$$$50$ million per $1$% more of the pollutant. It is too costly to remove all of the pollutant.
Solution

We first find $C'(x)\text{.}$

\begin{equation*} \begin{split} C'(x) \amp = \diff{ }{x} \left(\frac{0.5x}{100-x}\right)\\ \amp = 0.5\frac{(100-x)-x(-1)}{(100-x)^{2}} \\ \amp = \frac{50}{(100-x)^{2}} \cdot \end{split} \end{equation*}

Therefore, $C'(80) = \dfrac{50}{20^{2}} = 0.125\text{,}$ $C'(90) = \dfrac{50}{10^{2}} = 0.5\text{,}$ $C'(95) = \dfrac{50}{5^{2}} = 2$ and $C'(99) = \dfrac{50}{1} = 50\text{.}$ I.e., the cost to remove $80$% of the toxic waste is $$0.125$ million, to remove $90$% of the waste is$$0.5$ million, etc. Notice that, as $x$ gets closer and closer to $100$%, the cost to remove the pollutant is increasing at an increasingly faster rate. In particular, we see that it would cost approximately $$50$ million dollars to remove just over $99$% of the waste. The total worldwide box-office receipts for a long-running movie are approximated by the function \begin{equation*} T(x)=\frac{120x^{2}}{x^{2}+4} \end{equation*} where $T(x)$ is measured in millions of dollars and $x$ is the number of years since the movie's release. How fast are the total receipts changing $1\text{,}$ $3$ and $5$ years after its release? Answer$$38.4\text{,}$ $$17.04\text{,}$$$5.71$ million per year.

Grand & Toy makes a line of executive desks. It is estimated that the total cost for making $x$ units of their Senior Executive model is

\begin{equation*} C(x)=100x+200,000 \end{equation*}

dollars/year

1. Find the average cost function, $\overline{C}\text{.}$

Solution
$\displaystyle{\overline{C}(x) = \frac{C(x)}{x} = \frac{100x+200,000}{x} = 100 + \frac{200,000}{x}.}$
2. Find the marginal average cost function, $\overline{C}'\text{.}$

Solution
$\displaystyle{\overline{C'}(x) = \diff{}{x} \left(100 + \frac{200,000}{x}\right) = -\frac{200,000}{x^2}.}$
3. What happens to $\overline{C}(x)$ when $x$ is very large? Interpret your results.

Solution

We compute

\begin{equation*} \lim_{x\to\infty} \overline{C}(x) = \lim_{x\to\infty} \left(100+\frac{200,000}{x}\right) = 100\text{.} \end{equation*}

Therefore, if the production levels are very high, then the average cost of the desks approaches $100 per year per unit. The quantity of Sicard wristwatches demanded each month is related to the unit price by the equation \begin{equation*} p= \frac{50}{0.001x^{2}+1} \ \ \ 0 \leq x \leq 20 \end{equation*} where $p$ is measured in dollars and $x$ in units of a thousand. 1. Find the revenue function $R\text{.}$ Solution $R(x) = px = \dfrac{50x}{0.01x^{2}+1}\text{.}$ 2. Find the marginal revenue function $R'\text{.}$ Solution $R'(x) = \dfrac{0.01x^{2}+1)(50)-(50x)(0.02x)}{(0.01x^{2}+1)^{2}} = \dfrac{50-0.5x^{2}}{(0.01x^{2}+1)^{2}} \cdot$ 3. Compute $R'(2)$ and interpret your result. Solution $R'(2) = \dfrac{50-0.5(4)}{(0.01(4)+1)^{2}} \approx 44.379\text{.}$ We see that, when sales are at a level of $2000$ units, the revenue is increasing at a rate of approximately$$44,379$ per $1000$ units.