
## Section2.1Elementary Derivative Rules

###### Motivating Questions
• What are alternative notations for the derivative?

• How can we use the algebraic structure of a function $f(x)$ to compute a formula for $f'(x)\text{?}$

• What is the derivative of a power function of the form $f(x) = x^n\text{?}$ What is the derivative of an exponential function of the form $f(x) = a^x\text{?}$

• If we know the derivative of $y = f(x)\text{,}$ what is the derivative of $y = k f(x)\text{,}$ where $k$ is a constant?

• If we know the derivatives of $y = f(x)$ and $y = g(x)\text{,}$ how do we compute the derivative of $y = f(x) + g(x)\text{?}$

In Chapter1 we developed the concept of the derivative of a function. We now know that the derivative $f'$ of a function $f$ measures the instantaneous rate of change of $f$ with respect to $x\text{.}$ The derivative also tells us the slope of the tangent line to the graph of $y=f(x)$ at any given value of $x\text{.}$ So far, we have focused on interpreting the derivative graphically or, in the context of a physical setting, as a meaningful rate of change. To calculate the value of the derivative at a specific point, we have relied on the limit definition of the derivative:

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

In this chapter we investigate how the limit definition of the derivative leads to interesting patterns and rules that enable us to quickly find a formula for $f'(x)$ without directly using the limit definition. For example, we would like to apply shortcuts to differentiate a function such as $g(x) = 4x^7 - \sin(x) + 3e^x\text{.}$

###### Example2.1

Functions of the form $f(x) = x^n\text{,}$ where $n = 1, 2, 3, \ldots\text{,}$ are often called power functions. The first two questions below revisit work we did earlier in Chapter1, and the following questions extend those ideas to higher powers of $x\text{.}$

1. Use the limit definition of the derivative to find $F'(x)$ for $F(x) = x^2\text{.}$

2. Use the limit definition of the derivative to find $g'(x)$ for $g(x) = x^3\text{.}$

3. Use the limit definition of the derivative to find $G'(x)$ for $G(x) = x^4\text{.}$ (Hint: $(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\text{.}$ Apply this rule to $(x+h)^4$ within the limit definition.)

4. Based on your work in (a), (b), and (c), what do you conjecture is the derivative of $m(x) = x^5\text{?}$ Of $p(x) = x^{13}\text{?}$

5. Conjecture a formula for the derivative of $f(x) = x^n$ that holds for any positive integer $n\text{.}$ That is, given $f(x) = x^n$ where $n$ is a positive integer, what do you think is the formula for $f'(x)\text{?}$

Hint
1. $(a+b)^2=a^2+2ab+b^2\text{.}$
2. $(a+b)^3=a^3+3a^2b+3ab^2+b^3\text{.}$
3. $(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\text{.}$
4. Use your answers to (a), (b), and (c) to find a pattern in the relationship between the formula for the function and the formula for its derivative in each case.
5. What pattern did you use in (d) to find $m'(x)$ when $m(x)=x^5\text{?}$ What happens when you replace the $5$ with $n$?
1. $F'(x)=2x\text{.}$
2. $g'(x)=3x^2\text{.}$
3. $G'(x)=4x^3\text{.}$
4. $m'(x)=5x^4$ when $m(x)=x^5\text{;}$ when $p(x)=x^{13}\text{,}$ then $p'(x)=13x^{12}\text{.}$
5. $f'(x)=nx^{n-1}\text{,}$ where $f(x)=x^n$ and $n$ is a positive integer.
Solution
1. \begin{align*} F'(x) \amp = \lim_{h \to 0} \frac{F(x+h)-F(x)}{h}\\ \amp = \lim_{h\to0}\frac{(x+h)^2-x^2}{h}\\ \amp = \lim_{h\to0}\frac{(x^2+2xh+h^2)-x^2}{h}\\ \amp = \lim_{h\to0}\frac{2xh+h^2}{h}\\ \amp = \lim_{h\to0}(2x+2h)\\ \amp = 2x\text{.} \end{align*}
2. \begin{align*} g'(x) \amp = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}\\ \amp = \lim_{h\to0}\frac{(x+h)^3-x^3}{h}\\ \amp = \lim_{h\to0}\frac{(x^3+3x^2h+3xh^2+h^3)-x^3}{h}\\ \amp = \lim_{h\to0}\frac{3x^2h+3xh^2+h^3}{h}\\ \amp = \lim_{h\to0}(3x^2+3xh+h^2)\\ \amp = 3x^2\text{.} \end{align*}
3. \begin{align*} G'(x) \amp = \lim_{h \to 0} \frac{G(x+h)-G(x)}{h}\\ \amp = \lim_{h\to0}\frac{(x+h)^4-x^4}{h}\\ \amp = \lim_{h\to0}\frac{(x^4+4x^3h+6x^2h^2+4xh^3+h^4)-x^4}{h}\\ \amp = \lim_{h\to0}\frac{4x^3h+6x^2h^2+4xh^3+h^4}{h}\\ \amp = \lim_{h\to0}(4x^3+6x^2h+4xh^2+h^3)\\ \amp = 4x^3\text{.} \end{align*}
4. In the previous parts, we discovered that the derivatives of $x^2\text{,}$ $x^3\text{,}$ and $x^4$ are $2x\text{,}$ $3x^2\text{,}$ and $4x^3\text{,}$ respectively. It looks like when we increase the power of $x$ by one, this affects the derivative by increasing both the coefficient and the power of $x$ by one. For instance, going from $F(x)=x^2$ to $g(x)=x^3\text{,}$ we saw the derivatives go from $F'(x)=2x$ to $g'(x)=3x^2\text{.}$

Since the derivative of $x^4$ is $4x^3\text{,}$ if this pattern continues, then it makes sense for the derivative of $x^5$ to be $5x^4\text{.}$ Then if $p(x)=x^{13}\text{,}$ we've increased the power of $x$ by eight from $x^5\text{,}$ so we want to increase the coefficient and power in the derivative by eight as well, to get $p'(x)=(5+8)x^{4+8}=13x^{12}\text{.}$

5. The pattern we've seen is that increasing the power of $x$ by one makes the coefficient and power of $x$ in the derivative by one. Also, the power of $x$ in the derivative is one less than in the original function in each case we've looked at, and the coefficient on the derivative is the same as the power in the original function. For instance, $m(x)=x^5$ implies $m'(x)=5x^4=5x^{5-1}\text{.}$ If we try to generalize this by replacing $5$ with $n$, we conjecture that when $n$ is a positive integer and $f(x)=x^n\text{,}$ then $f'(x)=nx^{n-1}\text{.}$

### SubsectionSome Key Notation

In addition to our usual $f'$ notation, there are other ways to denote the derivative of a function, as well as the instruction to take the derivative. If we are thinking about the relationship between $y$ and $x\text{,}$ we sometimes denote the derivative of $y$ with respect to $x$ by the symbol

\begin{equation*} \frac{dy}{dx} \end{equation*}

which we read "dee-$y$ dee-$x\text{.}$" For example, if $y = x^2\text{,}$ we'll write that the derivative is $\frac{dy}{dx} = 2x\text{.}$ This notation comes from the fact that the derivative is related to the slope of a line, and slope is measured by $\frac{\Delta y}{\Delta x}\text{.}$ Note that while we read $\frac{\Delta y}{\Delta x}$ as change in $y$ over change in $x\text{,}$ we view $\frac{dy}{dx}$ as a single symbol, not a quotient of two quantities.

We use a variant of this notation as the instruction to take the derivative. In particular,

\begin{equation*} \frac{d}{dx}\left[ \Box \right] \end{equation*}

means take the derivative of the quantity in $\Box$ with respect to $x\text{.}$ For example, we may write $\frac{d}{dx}[x^2] = 2x\text{.}$

It is important to note that the independent variable can be different from $x\text{.}$ If we have $f(z) = z^2\text{,}$ we then write $f'(z) = 2z\text{.}$ Similarly, if $y = t^2\text{,}$ we say $\frac{dy}{dt} = 2t\text{.}$ And it is also true that $\frac{d}{dq}[q^2] = 2q\text{.}$ This notation may also be used for second derivatives: $f''(z) = \frac{d}{dz}\left[\frac{df}{dz}\right] = \frac{d^2 f}{dz^2}=\frac{d^2}{dz^2}[f]\text{.}$

In what follows, we'll build a repertoire of functions for which we can quickly compute the derivative.

### SubsectionConstant, Power, and Exponential Functions

So far, we know the derivative formula for two important classes of functions: constant functions and power functions. If $f(x) = c$ is a constant function, its graph is a horizontal line with slope zero at every point. Thus, $\frac{d}{dx}[c] = 0\text{.}$ We summarize this with the following rule.

###### Derivatives of Constant Functions

For any real number $c\text{,}$ if $f(x) = c\text{,}$ then $f'(x) = 0\text{.}$

###### Example2.2

If $f(x) = 7\text{,}$ then $f'(x) = 0\text{.}$ Similarly, $\frac{d}{dx} [\sqrt{3}] = 0\text{.}$

In your work in Example2.1, you conjectured that for any positive integer $n\text{,}$ if $f(x) = x^n\text{,}$ then $f'(x) = nx^{n-1}\text{.}$ This rule can be formally proved for any positive integer $n\text{,}$ and also for any nonzero real number (positive or negative).

###### Derivatives of Power Functions

For any nonzero real number $n\text{,}$ if $f(x) = x^n\text{,}$ then $f'(x) = nx^{n-1}\text{.}$

###### Example2.3

Using the rule for power functions, we can compute the following derivatives. If $g(z) = z^{-3}\text{,}$ then $g'(z) = -3z^{-4}\text{.}$ Similarly, if $h(t) = t^{\frac{7}{5}}\text{,}$ then $\frac{dh}{dt} = \frac{7}{5}t^{\frac{2}{5}}\text{,}$ and $\frac{d}{dq} [q^{\pi}] = \pi q^{\pi - 1}\text{.}$

It will be instructive to have a derivative formula for one more type of basic function. For now, we simply state this rule without explanation or justification. We will encounter graphical reasoning for why the rule is plausible in Example2.6, and we will further explore why this rule is true in one of the exercises.

###### Derivatives of Exponential Functions

For any positive real number $a\text{,}$ if $f(x) = a^x\text{,}$ then $f'(x) = a^x \ln(a)\text{.}$

###### Example2.4

If $f(x) = 2^x\text{,}$ then $f'(x) = 2^x \ln(2)\text{.}$ Similarly, for $p(t) = 10^t\text{,}$ $p'(t) = 10^t \ln(10)\text{.}$ It is especially important to note that when $a = e\text{,}$ where $e$ is the base of the natural logarithm function, we have that

\begin{equation*} \frac{d}{dx} [e^x] = e^x \ln(e) = e^x \end{equation*}

since $\ln(e) = 1\text{.}$ This is an extremely important property of the function $e^x\text{:}$ it is its own derivative!

Note carefully the distinction between power functions and exponential functions: in power functions, the variable is in the base, as in $x^2\text{,}$ while in exponential functions, the variable is in the exponent, as in $2^x\text{.}$ As we can see from the rules, this makes a big difference in the form of the derivative.

###### Example2.5

Use the three rules above to determine the derivative of each of the following functions. For each, state your answer using full and proper notation, labeling the derivative with its name. For example, if you are given a function $h(z)\text{,}$ you should write $h'(z) =$ or $\frac{dh}{dz} =$ as part of your response.

1. $f(t) = \pi$

2. $g(z) = 7^z$

3. $h(w) = w^{\frac{3}{4}}$

4. $p(x) = 3^{\frac{1}{2}}$

5. $r(t) = (\sqrt{2})^t$

6. $s(q) = q^{-1}$

7. $m(t) = \frac{1}{t^3}$

Hint
1. Is $\pi$ a variable or a constant?

2. Is $g$ a power function or an exponential function?

3. Is $h$ a power function or an exponential function?

4. Is $3^{\frac{1}{2}}$ a constant or a variable expression?

5. $\sqrt{2}$ is a constant.

6. Is $s$ a power function or an exponential function?

7. Try rewriting the function using a negative exponent instead of a fraction.

1. $f'(t) = 0\text{.}$

2. $g'(z) = 7^z \ln(7)\text{.}$

3. $h'(w) = \frac{3}{4} w^{\frac{-1}{4}}\text{.}$

4. $\frac{dp}{dx} = 0\text{.}$

5. $r'(t) = (\sqrt{2})^t \ln (\sqrt{2})\text{.}$

6. $\frac{d}{dq}[q^{-1}] = -q^{-2}\text{.}$

7. $\frac{dm}{dt} = -3t^{-4} = -\frac{3}{t^4}\text{.}$

Solution
1. $f(t) = \pi$ is constant, so $f'(t) = 0\text{.}$

2. $g(z) = 7^z$ is an exponential function with base $7\text{,}$ so $g'(z) = 7^z \ln(7)\text{.}$

3. $h(w) = w^{\frac{3}{4}}$ is a power function, thus $h'(w) = \frac{3}{4} w^{(\frac{3}{4})-1}=\frac{3}{4} w^{\frac{-1}{4}}\text{.}$

4. $p(x) = 3^{\frac{1}{2}}=\sqrt{3}$ is constant, and therefore $\frac{dp}{dx} = 0\text{.}$

5. $r(t) = (\sqrt{2})^t$ is exponential with base $\sqrt{2}$ (since $\sqrt{2}$ is a constant), and so we have $r'(t) = (\sqrt{2})^t \ln (\sqrt{2})\text{.}$

6. $s(q)=q^{-1}$ is a power function, so $s'(q)=(-1)q^{(-1)-1}=-q^{-2}\text{.}$

7. $m(t) = \frac{1}{t^3} = t^{-3}$ is a power function, so $\frac{dm}{dt} = -3t^{-4} = -\frac{3}{t^4}\text{.}$

The goal of this next example is to provide insight into why the derivative rule for exponential functions is true. We will investigate the plausibility of the rule through exploration of the graphs of an exponential function and its derivative.

###### Example2.6

Consider the function $g(x) = 2^x\text{,}$ which is graphed below in Figure2.7.

1. At each of $x = -2, -1, 0, 1, 2\text{,}$ use a straightedge to sketch an accurate tangent line to $y = g(x)\text{.}$

2. For each tangent line you drew in (a), use the provided grid to estimate the slope.

3. Use the limit definition of the derivative to estimate $g'(0)$ by using small values of $h\text{,}$ and compare the result to your visual estimate for the slope of the tangent line to $y = g(x)$ at $x = 0$ in (b).

4. Based on your work in (a), (b), and (c), sketch an accurate graph of $y = g'(x)$ on the interval $-3\le x\le 3\text{.}$

5. Write at least one sentence that explains why it is reasonable to think that $g'(x) = c\cdot g(x)\text{,}$ where $c$ is a constant. In addition, calculate $\ln(2)\text{,}$ and then discuss how this value, combined with your work above, reasonably suggests that $g'(x) = 2^x \ln(2)\text{.}$

Hint
1. The tangent line at $x=-2$ should be the flattest; the tangent line at $x=2$ should be the steepest.

2. Based on the axis labels, how big is each grid square?

3. It will be useful to remember that $2^0=1\text{.}$

4. How does the slope of a tangent line to $y=g(x)$ at a point $a$ relate to the value of $g'(a)\text{?}$

5. What sort of graph transformation occurs from the graph of $y=f(x)$ to the graph of $y=c\cdot f(x)$ for a given function $f$ and constant $c\text{?}$

1. The different tangent lines are graphed below; you can zoom in on any graph by clicking on it.

2. The slopes of the tangent lines to $g$ at $x=-2\text{,}$ $-1\text{,}$ $0\text{,}$ $1\text{,}$ and $2$ are approximately $0.17\text{,}$ $0.33\text{,}$ $0.71\text{,}$ $1.4\text{,}$ and $2.67\text{,}$ respectively.

3. $g'(0)\approx0.69\text{,}$ which matches the visual estimate obtained in (b).

4. Multiplying a function by a positive constant applies a vertical stretch to the graph.

Solution
1. The different tangent lines are graphed below in green; you can zoom in on any graph by clicking on it.

• Looking at the tangent line to $g$ at $x=-2\text{,}$ we see that on the interval $-3\le x\le 2$ the line rises by a little less than $1$ unit. This gives us a slope of approximately $0.8/5$ or $0.9/5\text{,}$ so we'll average those and say the slope of the tangent line to $y=g(x)$ at $x=-2$ is approximately $0.17\text{.}$

• Looking at the tangent line to $g$ at $x=-1\text{,}$ we see that on the interval $-1\le x\le 2$ the line rises by about $1$ unit. This gives us a slope of approximately $1/3\approx0.33\text{.}$
• Looking at the tangent line to $g$ at $x=0\text{,}$ we see that the line rises by $2$ units over the approximate interval $0\le x\le2.8\text{,}$ giving us a slope of about $2/2.8\approx0.71$ for this line.
• Looking at the tangent line to $g$ at $x=1\text{,}$ we see that the line rises by just over $4$ units on the interval $0\le x\le 3\text{.}$ This tells us the slope should be just over $4/3\text{,}$ possibly around $1.4\text{.}$
• Looking at the tangent line to $g$ at $x=2\text{,}$ we see that the line rises by $4$ units on about the interval $0.5\le x\le 2\text{,}$ so the slope of this line is approximately $4/1.5\approx2.67\text{.}$
2. The limit definition of $g'(0)$ says

\begin{equation*} g'(0)=\lim_{h\to0}\frac{g(0+h)-g(0)}{h}=\lim_{h\to0}\frac{2^h-2^0}{h}=\lim_{h\to0}\frac{2^h-1}{h}\text{.} \end{equation*}

We consider this difference quotient at the values $h=0.1\text{,}$ $h=0.01\text{,}$ and $h=0.001\text{,}$ and find that

\begin{align*} \frac{2^{0.1}-1}{0.1}\approx\mathstrut \amp 0.7177\\ \frac{2^{0.01}-1}{0.01}\approx\mathstrut \amp 0.6956\\ \frac{2^{0.001}-1}{0.001}\approx\mathstrut \amp 0.6933\text{.} \end{align*}

Hence we approximate $g'(0)$ by $0.693\text{,}$ which is very close to our slope estimate of $0.71$ from (b) for the tangent line to $y=g(x)$ at $x=0\text{.}$

3. We start with the $y$-intercept of $g'$ that we found in part (c): the point $(0,0.693)\text{.}$ Then, since the slope of the tangent line to $g$ at a point $x=a$ is exactly the value of $g'(a)\text{,}$ we use our slope estimates from (b) to add more points to our graph:

\begin{equation*} (-2,0.17),~ (-1,0.33),~ (1,1.4),~ (2,2.67)\text{.} \end{equation*}

Finally, since the graph of $y=g(x)$ is always increasing and concave up, we know the graph of $y=g'(x)$ will be always positive and increasing. Thus we end up with the sketch shown below.

4. Multiplying a function $g$ by a positive constant $c$ affects the graph of $y=g(x)$ by vertically stretching it if $1\lt c\text{,}$ or by vertically compressing it if $0\lt c\lt 1\text{.}$ The shapes of the graphs of $y=g(x)$ and $y=g'(x)$ are very similar, and it looks like the derivative graph is a slightly flattened version of the graph of $y=g(x)\text{.}$ This suggests $g'(x)=c\cdot g(x)$ for some constant $c$ between $0$ and $1\text{.}$ In particular, we note that $\ln(2)\approx0.693\text{.}$ Since this is the approximation we found for $g'(0)\text{,}$ where $g(0)=1\text{,}$ it is reasonable to assume that a formula for $g'$ is $g'(x)=\ln(2)\cdot g(x)=2^x\ln(2)\text{.}$

### SubsectionConstant Multiples and Sums of Functions

Next we will learn how to compute the derivative of a function constructed as an algebraic combination of basic functions. For instance, we'd like to be able to take the derivative of a polynomial function such as

\begin{equation*} p(t) = 3t^5 - 7t^4 + t^2 - 9\text{,} \end{equation*}

which is a sum of constant multiples of powers of $t\text{.}$ To that end, we develop two new rules: the constant multiple rule and the sum rule.

With the constant multiple rule, we want to answer the question how is the derivative of $y = kf(x)$ related to the derivative of $y = f(x)\text{?}$ Recall that when we multiply a function by a constant $k\text{,}$ we vertically stretch the graph by a factor of $|k|$ (and reflect the graph across the horizontal line $y = 0$ if $k \lt 0$). This vertical stretch affects the slope of the graph, so the slope of the function $y = kf(x)$ is $k$ times as steep as the slope of $y = f(x)\text{.}$ Thus, when we multiply a function by a factor of $k\text{,}$ we change the value of its derivative by a factor of $k$ as well.1We note that the constant multiple rule can be formally proved as a consequence of properties of limits, using the limit definition of the derivative.

###### The Constant Multiple Rule

For any real number $k\text{,}$ if $f(x)$ is a differentiable function with derivative $f'(x)\text{,}$ then

\begin{equation*} \frac{d}{dx}[k f(x)] = k f'(x)\text{.} \end{equation*}

In words, this rule says that the derivative of a constant times a function is the constant times the derivative of the function.

###### Example2.20

If $g(t) = 3 \cdot 5^t\text{,}$ then2Be careful with this one! $3\cdot 5^t$ is not the same as $15^t\text{.}$ An easy way to see the difference is to substitute $t=2$ into each expression: $g(2)=3\cdot 5^2=3\cdot25=75\text{;}$ whereas $15^2=225\text{.}$ $g'(t) = 3 \cdot 5^t \ln(5)\text{.}$ Similarly, $\frac{d}{dz} [5z^{-2}] = 5 (-2z^{-3})=-10z^{-3}\text{.}$

Next we examine a sum of two functions. If we have $y = f(x)$ and $y = g(x)\text{,}$ we can compute a new function $y = (f+g)(x)$ by adding the outputs of the two functions: $(f+g)(x) = f(x) + g(x)\text{.}$ Not only is the value of the new function the sum of the values of the two known functions, but the slope of the new function is the sum of the slopes of the known functions. Therefore,3Like the constant multiple rule, the sum rule can be formally proved as a consequence of properties of limits, using the limit definition of the derivative. we arrive at the following sum rule for derivatives:

###### The Sum Rule

If $f(x)$ and $g(x)$ are differentiable functions with derivatives $f'(x)$ and $g'(x)$ respectively, then

\begin{equation*} \frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\text{.} \end{equation*}

In words, the sum rule tells us that the derivative of a sum is the sum of the derivatives. It also tells us that a sum of two differentiable functions is also differentiable. Furthermore, because we can view the difference function $y = (f-g)(x) = f(x) - g(x)$ as $y = f(x) + (-1 \cdot g(x))\text{,}$ the sum rule and constant multiple rules together tell us that

\begin{equation*} \frac{d}{dx}\left[f(x)-g(x)\right]=\frac{d}{dx}\left[f(x) + (-1 \cdot g(x))\right] = f'(x) - g'(x)\text{,} \end{equation*}

or that the derivative of a difference is the difference of the derivatives. We can now compute derivatives of sums and differences of elementary functions.

###### Example2.21

Using the sum rule, we find that $\frac{d}{dw} (2^w + w^2) = 2^w \ln(2) + 2w\text{.}$ Using both the sum and constant multiple rules, if $h(q) = 3q^6 - 4q^{-3}\text{,}$ then $h'(q) = 3 (6q^5) - 4(-3q^{-4}) = 18q^5 + 12q^{-4}\text{.}$

###### Example2.22

Use only the rules for constant, power, and exponential functions, together with the constant multiple and sum rules, to compute the derivative of each function below with respect to the given independent variable. Note well that we do not yet know any rules for how to differentiate the product or quotient of functions. This means that you may have to first do some algebra on the functions below before you can actually use existing rules to compute the desired derivative formula. In each case, label the derivative you calculate with its name, using proper notation such as $f'(x)\text{,}$ $h'(z)\text{,}$ $\frac{dr}{dt}\text{,}$ etc.

1. $f(x) = x^{\frac{5}{3}} - x^4 + 2^x$

2. $g(x) = 14e^x + 3x^5 - x$

3. $h(z) = \sqrt{z} + \frac{1}{z^4} + 5^z$

4. $r(t) = \sqrt{53} \, t^7 - \pi e^t + e^4$

5. $s(y) = (y^2 + 1)(y^2 - 1)$

6. $p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12$

7. $q(x) = \frac{x^3 - x + 2}{x}$

Hint
1. Use the sum rule.

2. Use the sum rule together with the constant multiple rule.

3. How can you rewrite $\sqrt{z}$ using exponents?

4. Is $e^4$ a constant or variable expression?

5. Expand the product before attempting to find the derivative.

6. Note that $a$ is the independent variable.

7. Rewrite the single fraction as a sum of three fractions, and simplify.

1. $f'(x) = \frac{5}{3}x^{\frac{2}{3}} - 4 x^3 + 2^x \ln(2)\text{.}$

2. $g'(x) = 14e^x + 3 \cdot 5x^4 - 1\text{.}$

3. $h'(z) = \frac{1}{2}z^{-\frac{1}{2}} - 4z^{-5} + 5^z \ln(5)\text{.}$

4. $\frac{dr}{dt} = \sqrt{53} \cdot 7 t^6 - \pi e^t\text{.}$

5. $\frac{ds}{dy} = 4y^3\text{.}$

6. $p'(a) = 3\cdot4a^3 - 2\cdot3 a^2 + 7\cdot2a - 1\text{.}$

7. $q'(x) = 2x - 2x^{-2}\text{.}$

Solution
1. $f(x) = x^{\frac{5}{3}} - x^4 + 2^x\text{,}$ so by the sum (and difference) rule,

\begin{align*} f'(x) =\mathstrut \amp \frac{d}{dx}\left[x^{\frac{5}{3}}\right]-\frac{d}{dx}\left[x^4\right]+\frac{d}{dx}\left[2^x\right]\\ =\mathstrut \amp \frac53x^{(\frac{5}{3})-1}-4x^{4-1}+2^x\ln(2)\\ =\mathstrut \amp \frac53x^{\frac{2}{3}}-4x^3+2^x\ln(2)\text{.} \end{align*}
2. $g(x) = 14e^x + 3x^5 - x\text{,}$ so by the sum and constant multiple rules,

\begin{align*} g'(x) =\mathstrut \amp 14\frac{d}{dx}\left[e^x\right]+3\frac{d}{dx}\left[x^5\right]-\frac{d}{dx}[x]\\ =\mathstrut \amp 14(e^x)+3(5x^4)-1\\ =\mathstrut \amp 14e^x+15x^4-1\text{.} \end{align*}
3. $h(z) = \sqrt{z} + \frac{1}{z^4} + 5^z = z^{\frac{1}{2}} + z^{-4} + 5^z\text{,}$ thus

\begin{align*} h'(z) =\mathstrut \amp \frac{d}{dz}\left[z^{\frac{1}{2}}\right]+\frac{d}{dz}\left[z^{-4}\right]+\frac{d}{dz}\left[5^z\right]\\ =\mathstrut \amp \frac12z^{(\frac{1}{2})-1}-4z^{(-4)-1}+5^z\ln(5)\\ =\mathstrut \amp \frac12z^{-\frac{1}{2}}-4z^{-5}+5^z\ln(5)\text{.} \end{align*}
4. $r(t)=\sqrt{53}\,t^7-\pi e^t+e^4\text{,}$ so using the sum and constant multiple rules, we get

\begin{align*} r'(t) =\mathstrut \amp \sqrt{53}\,\frac{d}{dt}\left[t^7\right]-\pi\frac{d}{dt}\left[e^t\right]+\frac{d}{dt}\left[e^4\right]\\ =\mathstrut \amp \sqrt{53}\,(7t^6)-\pi(e^t)+0\\ =\mathstrut \amp 7\sqrt{53}\,t^6-\pi e^t\text{.} \end{align*}

(Note particularly that $\frac{d}{dt}[e^4] = 0$ since $e^4$ is constant.)

5. $s(y) = (y^2 + 1)(y^2 - 1)= y^4 - 1\text{,}$ thus $\frac{ds}{dy} = 4y^3\text{.}$

6. $p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12\text{,}$ so $p'(a) =3(4a^3)-2(3a^2)+7(2a)-1=12a^3 - 6 a^2 + 14a - 1\text{.}$

7. $q(x) = \frac{x^3 - x + 2}{x} = \frac{x^3}{x} - \frac{x}{x} + \frac{2}{x} = x^2 - 1 + 2x^{-1}\text{.}$ Using the power rule repeatedly, it follows that $q'(x) = 2x - 2x^{-2}\text{.}$

In the same way that we have shortcut rules to help us find derivatives, we introduce some language that is simpler and shorter. Often, rather than say take the derivative of $f\text{,}$ we'll instead say simply differentiate $f\text{.}$ Similarly, if the derivative exists at a point, we say $f$ is differentiable at that point, or that $f$ can be differentiated there.

As we work with the algebraic structure of functions, it is important to develop a big picture view of what we are doing. Here, we make several general observations based on the rules we have so far.

• The derivative of any polynomial function is another polynomial function, and the degree of the derivative is one less than the degree of the original function. For instance, if $p(t) = 7t^5 - 4t^3 + 8t\text{,}$ then $p$ is a degree 5 polynomial and its derivative, $p'(t) = 35t^4 - 12t^2 + 8\text{,}$ is a degree 4 polynomial.
• The derivative of any exponential function is another exponential function: for example, if $g(z) = 7 \cdot 2^z\text{,}$ then $g'(z) = 7 \cdot 2^z \ln(2)\text{,}$ which is also exponential.
• We should not lose sight of the fact that everything we developed in Chapter1 regarding the meaning of the derivative still holds. The derivative measures the instantaneous rate of change of the original function as well as the slope of the tangent line at any selected point on the curve.
###### Example2.23

Each of the following questions asks you to use derivatives to answer key questions about functions. Be sure to think carefully about each question and to use proper notation in your responses.

1. If $h(z)=\sqrt{z}+\frac1z\text{,}$ find the slope of the tangent line to the graph of $y=h(z)$ at the point where $z = 4\text{.}$

2. A population of cells is growing in such a way that its total number in millions is given by the function $P(t) = 2(1.37)^t + 32\text{,}$ where $t$ is measured in days.

1. Determine the instantaneous rate at which the population is growing on day 4, and include units in your answer.

2. Is the population growing at an increasing rate or growing at a decreasing rate on day 4? Explain.

3. Find an equation for the tangent line to the curve $y=p(a)\text{,}$ where $p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12\text{,}$ at the point where $a=-1\text{.}$

4. What is the difference between being asked to find the slope of the tangent line (asked in (a)) and the equation of the tangent line (asked in (c))?

Hint
1. How would $h'(z)$ help you answer the question?

2. Think about finding both $P'(t)$ and $P''(t)\text{.}$

3. Recall the equation of a tangent line from Section1.6.

4. What information do you find in both (a) and (c)? Which part requires more information? Why?

1. $h'(4) = \frac{3}{16}\text{.}$

1. $P'(4) = 2(1.37)^4 \ln(1.37) \approx 2.218$ million cells per day.

2. The population is growing at an increasing rate.

2. $y - 25 = -33(a+1)\text{.}$

3. The slope is a number and is only one piece of the equation, which contains more information.

Solution
1. Note that since $h(z) = z^{\frac{1}{2}} + z^{-1}\text{,}$ we have $h'(z) = \frac{1}{2}z^{-\frac{1}{2}} - z^{-2}=\frac{1}{2\sqrt{z}}-\frac{1}{z^2}\text{.}$ Thus, $h'(4) = \frac{1}{2\sqrt4} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}\text{.}$ Therefore the slope of the tangent line to the graph of $y=h(z)$ at the point where $z = 4$ is $\frac{3}{16}\text{.}$

1. Note that $P'(t) = 2(1.37)^t \ln(1.37)\text{,}$ and therefore $P'(4) = 2(1.37)^4 \ln(1.37) \approx 2.218$ million cells per day.

2. We can compute $P''(t)$ by taking the derivative of $P'(t)\text{,}$ and we find that $P''(t) = 2(1.37)^t \ln(1.37) \ln(1.37)\text{.}$ Then we see that $P''(4) \approx 0.69825$ is positive, which tells us that $P'(t)$ is increasing at $t = 4\text{.}$ Since the instantaneous rate of change of $P$ is (positive and) increasing, it follows that the population is growing at an increasing rate on the fourth day.

2. Given $p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12\text{,}$ first observe that $p(-1) = 3 + 2 + 7 + 1 + 12 = 25\text{,}$ so the tangent line will pass through the point $(-1,25)\text{.}$ Furthermore, since $p'(a) = 12a^3 - 6a^2 + 14a - 1\text{,}$ we know that the slope of the tangent line is $p'(-1) = -12 - 6 - 14 - 1 = -33\text{.}$ An equation of the tangent line is therefore $y - 25 = -33(a+1)\text{.}$ Equivalently, $y=-33a-8$ is an equation of this tangent line.

3. Finding the slope of the tangent line only requires knowing the value of the derivative at a particular input. In contrast, finding the equation of the tangent line additionally requires knowing the value of the original function at that same input. In other words, the slope is just a number, whereas the equation incorporates additional information that together with the slope is sufficient to actually graph the tangent line.

### SubsectionSummary

• Given a differentiable function $y = f(x)\text{,}$ we can express the derivative of $f$ in several different notations: $f'(x)\text{,}$ $\frac{df}{dx}\text{,}$ $\frac{dy}{dx}\text{,}$ and $\frac{d}{dx}[f(x)]\text{.}$

• The limit definition of the derivative leads to patterns among certain families of functions. These patterns enable us to compute derivative formulas without resorting directly to the definition. For example, if $f$ is a power function of the form $f(x) = x^n$ for any real number $n$ other than 0, then $f'(x) = nx^{n-1}\text{.}$ This is called the rule for power functions.

• We have stated a rule for derivatives of exponential functions in the same spirit as the rule for power functions: for any positive real number $a\text{,}$ if $f(x) = a^x\text{,}$ then $f'(x) = a^x \ln(a)\text{.}$ The graphical exploration in Example2.6 provides visual evidence for this rule as it shows that for an exponential function $f(x)=a^x$ $(a \gt 1)\text{,}$ the graph of $y=f'(x)$ appears to be a scaled version of the original function. In particular, careful analysis of the graph of $y=2^x$ suggests that $\frac{d}{dx}[2^x]=2^x\ln(2)\text{.}$

• If we are given a constant multiple of a function whose derivative we know, or a sum of functions whose derivatives we know, the constant multiple and sum rules make it straightforward to compute the derivative of the overall function. More formally, if $f(x)$ and $g(x)$ are differentiable with derivatives $f'(x)$ and $g'(x)$ and if $a$ and $b$ are constants, then

\begin{equation*} \frac{d}{dx} \left[af(x) + bg(x)\right] = af'(x) + bg'(x)\text{.} \end{equation*}

### SubsectionExercises

Let $f$ and $g$ be differentiable functions for which the following information is known: $f(2) = 5\text{,}$ $g(2) = -3\text{,}$ $f'(2) = -1/2\text{,}$ $g'(2) = 2\text{.}$

1. Let $h$ be the new function defined by the rule $h(x) = 3f(x) - 4g(x)\text{.}$ Determine $h(2)$ and $h'(2)\text{.}$

2. Find an equation for the tangent line to $y = h(x)$ at the point $(2,h(2))\text{.}$

3. Let $p$ be the function defined by the rule $p(x) = -2f(x) + \frac{1}{2}g(x)\text{.}$ Is $p$ increasing, decreasing, or neither at $a = 2\text{?}$ Why?

4. Estimate the value of $p(2.03)$ by using the local linearization of $p$ at the point $(2,p(2))\text{.}$

Let functions $p$ and $q$ be the piecewise linear functions given by their respective graphs in Figure2.24. Use the graphs to answer the following questions.

1. At what values of $x$ is $p$ not differentiable? At what values of $x$ is $q$ not differentiable? Why?

2. Let $r(x) = p(x) + 2q(x)\text{.}$ At what values of $x$ is $r$ not differentiable? Why?

3. Determine $r'(-2)$ and $r'(0)\text{.}$

4. Find an equation for the tangent line to $y = r(x)$ at the point $(2,r(2))\text{.}$

Consider the functions $r(t) = t^t$ and $s(t) = \arccos(t)\text{,}$ for which you are given the facts that $r'(t) = t^t(\ln(t) + 1)$ and $s'(t) = -\frac{1}{\sqrt{1-t^2}}\text{.}$ Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain $0 \lt t \lt 1\text{.}$

1. Let $w(t) = 3t^t - 2\arccos(t)\text{.}$ Determine $w'(t)\text{.}$

2. Find an equation for the tangent line to $y = w(t)$ at the point $(\frac{1}{2}, w(\frac{1}{2}))\text{.}$

3. Let $v(t) = t^t + \arccos(t)\text{.}$ Is $v$ increasing or decreasing at the instant $t = \frac{1}{2}\text{?}$ Why?

Let $f(x) = a^x\text{.}$ The goal of this problem is to explore how the value of $a$ affects the derivative of $f(x)\text{,}$ without assuming we know the rule for $\frac{d}{dx}[a^x]$ that we have stated and used in earlier work in this section.

1. Use the limit definition of the derivative to show that

\begin{equation*} f'(x) = \lim_{h \to 0} \frac{a^x \cdot a^h - a^x}{h}\text{.} \end{equation*}
2. Explain why it is also true that

\begin{equation*} f'(x) = a^x \cdot \lim_{h \to 0} \frac{a^h - 1}{h}\text{.} \end{equation*}
3. Use computing technology and small values of $h$ to estimate the value of

\begin{equation*} L = \lim_{h \to 0} \frac{a^h - 1}{h} \end{equation*}

when $a = 2\text{.}$ Do likewise when $a = 3\text{.}$

4. Note that it would be ideal if the value of the limit $L$ was $1\text{,}$ for then $f$ would be a particularly special function: its derivative would be simply $a^x\text{,}$ which would mean that its derivative is itself. By experimenting with different values of $a$ between $2$ and $3\text{,}$ try to find a value for $a$ for which

\begin{equation*} L = \lim_{h \to 0} \frac{a^h - 1}{h} = 1\text{.} \end{equation*}
5. Compute $\ln(2)$ and $\ln(3)\text{.}$ What does your work in (b) and (c) suggest is true about $\frac{d}{dx}[2^x]$ and $\frac{d}{dx}[3^x]\text{?}$

6. How do your investigations in (d) lead to a particularly important fact about the function $f(x) = e^x\text{?}$