Section 2.3: The Power and Sum Rules for Derivatives

Example 1

Find the derivative of \( y=f(x)=mx+b \)

This is a linear function, so its graph is its own tangent line! The slope of the tangent line, the derivative, is the slope of the line: \[f'(x)=m\]

Example 2

Find the derivative of \( f(x)=135 \).

Think about this one graphically, too. The graph of f(x) is a horizontal line. So its slope is zero: \[f'(x)=0\]

Example 3

Find the derivative of \( f(x)=x^2 \).

Recall the formal definition of the derivative: \[f'(x)=\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}.\]

Using our function \( f(x)=x^2 \), \( f(x+h)=(x+h)^2=x^2+2xh+h^2 \).

Then \[ \begin{align*} f'(x) & = \lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}\\ & = \lim\limits_{h\to 0} \frac{x^2+2xh+h^2-x^2}{h}\\ & = \lim\limits_{h\to 0} \frac{2xh+h^2}{h}\\ & = \lim\limits_{h\to 0} \frac{h(2x+h)}{h}\\ & = \lim\limits_{h\to 0} (2x+h)\\ & = 2x \end{align*} \]

From all that, we find that \( f'(x)=2x \).

Example 4

Find the derivative of \( g(x)=4x^3 \)

Using the power rule, we know that if \( f(x)=x^3 \), then \( f'(x)=3x^2 \). Notice that \(g\) is 4 times the function \(f\). Think about what this change means to the graph of \(g\) – it’s now 4 times as tall as the graph of \(f\). If we find the slope of a secant line, it will be \( \frac{\Delta g}{\Delta x}= \frac{4\Delta f}{\Delta x} =4\frac{\Delta f}{\Delta x} \); each slope will be 4 times the slope of the secant line on the \(f\) graph. This property will hold for the slopes of tangent lines, too: \[\frac{d}{dx}\left(4x^3\right)=4\frac{d}{dx}\left(x^3\right)=4\cdot 3x^2=12x^2.\]

Example 5

Find the derivative of \( p(x)=17x^{10}+13x^8-1.8x+1003 \).

\[ \begin{align*} \frac{d}{dx}\left( 17x^{10}+13x^8-1.8x+1003 \right) & = \frac{d}{dx}\left( 17x^{10} \right)+\frac{d}{dx}\left( 13x^8 \right)-\frac{d}{dx}\left( 1.8x \right)+\frac{d}{dx}\left( 1003 \right)\\ & = 17\frac{d}{dx}\left( x^{10} \right)+13\frac{d}{dx}\left( x^8 \right)-1.8\frac{d}{dx}\left( x \right)+\frac{d}{dx}\left( 1003 \right)\\ & = 17\left(10x^9\right)+13\left(8x^7\right)-1.8\left(1\right)+0\\ & = 170x^9+104x^7-1.8 \end{align*} \]

Example 6

Find \(\frac{d}{dx}\left( 17x^2-33x+12 \right)\).

Writing out the rules, we'd write \[\frac{d}{dx}\left( 17x^2-33x+12 \right)=17(2x)-33(1)+0=34x-33.\]

Once you're familiar with the rules, you can, in your head, multiply the 2 times the 17 and the 33 times 1, and just write \[\frac{d}{dx}\left( 17x^2-33x+12 \right)=34x-33.\]

Example 7

Find the derivative of \( y=3\sqrt{t}-\frac{4}{t^4}+5e^t \).

The first step is translate into exponents: \[y=3\sqrt{t}-\frac{4}{t^4}+5e^t=3t^{1/2}-4t^{-4}+5e^t\]

Now you can take the derivative: \[ \begin{align*} \frac{d}{dt}\left( 3t^{1/2}-4t^{-4}+5e^t \right) & = 3\left(\frac{1}{2}t^{-1/2}\right)-4\left(-4t^{-5}\right)+5\left(e^t\right) \\ & = \frac{3}{2}t^{-1/2}+16t^{-5}+5e^t \end{align*} \]

If there is a reason to, you can rewrite the answer with radicals and positive exponents: \[y'= \frac{3}{2}t^{-1/2}+16t^{-5}+5e^t= \frac{3}{2\sqrt{t}}+\frac{16}{t^5}+5e^t\]

Example 8

Find the equation of the line tangent to \( g(t)=10-t^2 \) when \(t = 2\).

The slope of the tangent line is the value of the derivative. We can compute \( g'(t)=-2t \). To find the slope of the tangent line when \(t = 2\), evaluate the derivative at that point. The slope of the tangent line is -4.

To find the equation of the tangent line, we also need a point on the tangent line. Since the tangent line touches the original function at \(t = 2\), we can find the point by evaluating the original function: \( g(2)=10-2^2=6 \). The tangent line must pass through the point (2, 6).

Using the point-slope equation of a line, the tangent line will have equation \( y-6=-4(t-2) \). Simplifying to slope-intercept form, the equation is \( y=-4t+14 \).

Graphing, we can verify this line is indeed tangent to the curve:

Example 9

In a memory experiment, a researcher asks the subject to memorize as many words from a list as possible in 10 seconds. Recall is tested, then the subject is given 10 more seconds to study, and so on. Suppose the number of words remembered after \(t\) seconds of studying could be modeled by \( W(t)=4t^{2/5} \). Find and interpret \( W'(20) \).

\( W'(t)=4\cdot \frac{2}{5}t^{-3/5}=\frac{8}{5}t^{-3/5} \), so \( W'(20)=\frac{8}{5}(20)^{-3/5}\approx 0.2652 \).

Since \(W\) is measured in words, and \(t\) is in seconds, \(W'\) has units words per second. \( W'(20)\approx 0.2652 \) means that after 20 seconds of studying, the subject is learning about 0.27 more words for each additional second of studying.

Example 10

The table shows the total cost (TC) of producing \(q\) items.

1. What is the fixed cost?
2. When 200 items are made, what is the total variable cost? The average variable cost?
3. When 200 items are made, estimate the marginal cost.

1. The fixed cost is $20,000, the cost even when no items are made.

2. When 200 items are made, the total cost is $45,000. Subtracting the fixed cost, the total variable cost is $45,000 - $20,000 = $25,000.

   The average variable cost is the total variable cost divided by the number of items, so we would divide the $25,000 total variable cost by the 200 items made. $25,000/200 = $125. On average, each item had a variable cost of $125.

3. We need to estimate the value of the derivative, or the slope of the tangent line at \(q = 200\). Finding the secant line from \(q=100\) to \(q=200\) gives a slope of \[ \frac{45,000-35,000}{200-100}=100.\]

   Finding the secant line from \(q=200\) to \(q=300\) gives a slope of \[\frac{53,000-45,000}{300-200}=80.\]

   We could estimate the tangent slope by averaging these secant slopes, giving us an estimate of $90/item.

   This tells us that after 200 items have been made, it will cost about $90 to make one more item.

Example 11

The cost to produce \(x\) items is \(C(x) = \sqrt{x}\) hundred dollars.

1. What is the cost for producing 100 items? 101 items? What is cost of the 101st item?
2. Calculate \(C '(x)\) and evaluate \(C '\) at \(x = 100\). How does \(C '(100)\) compare with the last answer in Part a?

1. \(C(100) =\) 10 hundred dollars = $1000 and \(C(101) =\)10.0499 hundred dollars = $1004.99, so it costs $4.99 for that 101st item. Using this definition, the marginal cost is $4.99.
2. \( C'(x)=\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\), so \( C'(100)=\frac{1}{2\sqrt{100}}=\frac{1}{20} \) hundred dollars = $5.00.

Note how close these answers are! This shows (again) why it’s OK that we use both definitions for marginal cost.

Example 12

The demand, \(D\), for a product at a price of \(p\) dollars is given by \( D(p)=200-0.2p^2 \). Find the marginal revenue when the price is $10.

First we need to form a revenue equation. Since Revenue = Price\( \times \)Quantity, and the demand equation shows the quantity of product that can be sold, we have \[R(p)=D(p)\cdot p=\left(200-0.2p^2\right)p=200p-0.2p^3.\]

Now we can find marginal revenue by finding the derivative: \[R'(p)=200(1)-0.2(3p^2)=200-0.6p^2\]

At a price of $10, \( R'(10)=200-0.6(10)^2=140 \).

Notice the units for \(R'\) are \(\frac{\text{dollars of Revenue}}{\text{dollar of price}}\), so \( R'(10)=140 \) means that when the price is $10, the revenue will increase by $140 for each dollar that the price was increased.