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## Section2.4Derivative 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 $f(x) = x^n\text{?}$

• 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 $y = f(x) + g(x)\text{?}$

Previously, in Section 2.2, we defined the derivative function as:

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

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.

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.

### SubsectionSome Key Notation

In addition to our usual $f'$ notation, there are other ways to denote the derivative of a function. 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 $\displaystyle \frac{dy}{dx} = 2x\text{.}$ 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 $\displaystyle \frac{dy}{dx}$ as a single symbol, not a quotient of two quantities. It is important to note that this is notation, that is $\displaystyle f'(x)=\frac{dy}{dx}$ both mean the same thing!

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 $\displaystyle f'(z) =\frac{df}{dz}= 2z\text{.}$ Similarly, if $y = t^2\text{,}$ we say $\displaystyle \frac{dy}{dt} = 2t\text{.}$ And it is also true that $\frac{d}{dq}[q^2] = 2q\text{.}$

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

### SubsectionConstant and Power 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 $f'(x) = 0\text{.}$

###### Example2.34

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

Previously we noted that the derivative of $f(x)=x^2$ is $f'(x)=2x\text{.}$ We derived this from our limit definition of the derivative. If we use the limit definition to find the derivative of $g(x)=x^3$ we have:

\begin{align*} g'(x)= \mathstrut \amp \lim_{h\to0}\frac{(x+h)^3-x^3}{h}\\ =\mathstrut \amp \lim_{h\to 0} \frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}\\ =\mathstrut \amp \lim_{h\to 0}\frac{3x^2h+3xh^2+h^3}{h}\\ =\mathstrut \amp \lim_{h\to 0} 3x^2+3xh+h^2=3x^2 \end{align*}

Thus, the derivative of $g(x)=x^3$ is $g'(x)=3x^2\text{.}$

If we use the limit definition to find the derivative of $h(x)=x^4$ we have:

\begin{align*} h'(x)=\mathstrut \amp \lim_{h\to0}\frac{(x+h)^4-x^4}{h}\\ =\mathstrut \amp \lim_{h\to 0} \frac{x^4+4x^3h+6x^2h^2+4xh^3+h^4-x^4}{h}\\ =\mathstrut \amp \lim_{h\to 0}\frac{4x^3h+6x^2h^2+4xh^3+h^4}{h}\\ =\mathstrut \amp \lim_{h\to 0} 4x^3+6x^2h+4xh^2+h^3=4x^3 \end{align*}

Thus, the derivative of $h(x)=x^4$ is $g'(x)=4x^3\text{.}$ We note a pattern here, where the power "comes down" and new power is one less then the old. We now generalize this.

###### Derivatives of Power Functions - Power Rule

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

###### Example2.35

Use the rule for power functions to compute the derivatives of the following functions.

1. $f(x)=x^8$

2. $h(x)=x^{4.2}$

3. $g(x)=\sqrt{x}=x^{1/2}$

Using the rule for power functions, we can compute the following derivatives.

1. $f'(x)=8x^7$

2. $h'(x)=4.2x^{3.2}$

3. $\displaystyle g'(x)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$

For part c we use the fact that the square root can be rewritten as a $\displaystyle \frac{1}{2}$ power.

###### Example2.36

Use the rule for power functions to compute the derivatives of the following functions.

1. $\displaystyle g(z)=\frac{1}{z^3}=z^{-3}$

2. $\displaystyle h(t)=t^{7/5}$

3. $\displaystyle f(q)=q^{\pi}$

Using the rule for power functions, we can compute the following derivatives.

1. $\displaystyle g'(z)=-3z^{-4}$

2. $\displaystyle \frac{dh}{dt}=\frac{7}{5}t^{2/5}$

3. $\displaystyle \frac{d}{dq}[q^{\pi}]=\pi q^{\pi-1}$

###### Example2.37

Use the two 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 $\displaystyle \frac{dh}{dz} =$ as part of your response.

1. $f(t) = \pi$

2. $\displaystyle g(z) = \frac{1}{\sqrt{z}}$

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

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

5. $\displaystyle r(t) = (\sqrt{t})^3$

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

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

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

2. $\displaystyle g'(z) = \frac{-1}{2z^{3/2}}\text{.}$

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

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

5. $\displaystyle r'(t) = \frac{3}{2}t^{1/2}\text{.}$

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

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

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

2. First rewrite as $\displaystyle g(z) = \frac{1}{z^{1/2}}=z^{-1/2}$ then apply the power rule $\displaystyle g'(z) = \frac{-1}{2}z^{-3/2}=\frac{-1}{2z^{3/2}}\text{.}$

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

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

5. First rewrite as $\displaystyle r(t) = t^{3/2}$ then apply the power rule $\displaystyle r'(t) = \frac{3}{2}t^{1/2}\text{.}$

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

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

### 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.

###### Example2.38

If $g(t) =6t^3\text{,}$ then

\begin{equation*} g'(t) = 6(3t^2)=18t^2\text{.} \end{equation*}

Similarly,

\begin{equation*} \displaystyle \frac{d}{dz} [5z^{-2}] = 5 (-2z^{-3})=-10z^{-3}\text{.} \end{equation*}

In practice we normally multiply the constant and the power in one step, for example if $f(x)=5x^4$ then $f'(x)=20x^3\text{.}$

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,2Like 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.

###### Example2.39

Using the sum rule, we find that

\begin{equation*} \frac{d}{dw} (w^4+w^7) = 4w^3+7w^6\text{.} \end{equation*}

Using both the sum and constant multiple rules, if $h(q) = 3q^6 - 4q^{-3}\text{,}$ then

\begin{equation*} h'(q) = 3 (6q^5) - 4(-3q^{-4}) = 18q^5 + 12q^{-4}\text{.} \end{equation*}
###### Example2.40

Use only the rules for constant and power 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 $f'(x)\text{,}$ $h'(z)\text{,}$ $\displaystyle \frac{dr}{dt}\text{,}$ etc.

1. $\displaystyle f(x) = x^{\frac{5}{3}} - x^4 + 20$

2. $g(x) = 14x^6 + 3x^5 - x$

3. $\displaystyle h(z) = \sqrt{z} + \frac{1}{z^4} + 50$

4. $\displaystyle s(y) = (y^2 + 1)(y^2 - 1)$

5. $\displaystyle p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12$

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

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

2. $\displaystyle g'(x) = 84x^5 + 3 \cdot 5x^4 - 1\text{.}$

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

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

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

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

Solution
1. $\displaystyle f(x) = x^{\frac{5}{3}} - x^4 + 20\text{,}$ 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[20\right]\\ =\mathstrut \amp \frac53x^{(\frac{5}{3})-1}-4x^{4-1}+0\\ =\mathstrut \amp \frac53x^{\frac{2}{3}}-4x^3\text{.} \end{align*}
2. $g(x) = 14x^6 + 3x^5 - x\text{,}$ so by the sum and constant multiple rules,

\begin{align*} g'(x) =\mathstrut \amp 14\frac{d}{dx}\left[x^6\right]+3\frac{d}{dx}\left[x^5\right]-\frac{d}{dx}[x]\\ =\mathstrut \amp 14(6x^5)+3(5x^4)-1\\ =\mathstrut \amp 84x^5+15x^4-1\text{.} \end{align*}
3. $\displaystyle h(z) = \sqrt{z} + \frac{1}{z^4} + 5^z = z^{\frac{1}{2}} + z^{-4} + 50\text{,}$ 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[50\right]\\ =\mathstrut \amp \frac12z^{(\frac{1}{2})-1}-4z^{(-4)-1}+0\\ =\mathstrut \amp \frac12z^{-\frac{1}{2}}-4z^{-5}\text{.} \end{align*}
4. $\displaystyle s(y) = (y^2 + 1)(y^2 - 1)= y^4 - 1\text{,}$ thus $\displaystyle \frac{ds}{dy} = 4y^3\text{.}$

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

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

In the same way that we have 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.
• We should not lose sight of the fact that everything we developed in Chapter 1 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.

### SubsectionSummary

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

• 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.

• 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*}