
## Section8.5Differentiation

The most important application of limit rules is to develop rules for derivatives. Every time we need a derivative, we currently must use the definition and compute the limit

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

and then go through the algebra and simplification to find the resulting formula. Finding the formula for $f'(x)$ from the formula for $f(x)$ is a process called differentiation. The rules for derivatives will provide us with a methodical way to differentiate algebraic formulas.

Differentiation is a process of taking a function and using it to determine another function. In a sense, we are using a function as an input and creating a new function as the output. Can you see that this is something like a function of functions instead of a function of numbers. We call such a process an operator and most commonly use the differential operator $\displaystyle \frac{d}{dx}$ where $x$ is the independent variable for the function. (The variable changes depending on the relevant independent variable.)

###### Definition8.5.1
The differential operator $\displaystyle \frac{d}{dx}$ takes a function as its input and provides the derivative function as its output,
\begin{equation*} \frac{d}{dx}[f(x)] = f'(x). \end{equation*}
If $y$ is a dependent variable defined by a function $y=f(x)\text{,}$ then we can also write
\begin{equation*} \frac{dy}{dx} = \frac{d}{dx}[y] = f'(x). \end{equation*}

### Subsection8.5.1Derivative Rules

Derivative rules are theorems that take as a hypothesis that one or two functions have known derivatives and the conclusion tells how to find the derivative of some combination of those functions. We start by stating the basic rules together for convenience in finding them.

One of these differentiation rules, the chain rule, will require its own discussion. That rule is focused on how to differentiate compositions of functions. The other rules focus on arithmetic combinations of functions and are the primary focus of this section. The chain rule was included for completeness in the listing of differentiation rules.

The proofs for these differentiation rules are based on applying the definition of a derivative to the formula in question while knowing that the limits that define the derivatives in the hypothesis are valid. To illustrate, we will look at four of the differentiation rules in detail. Before doing this, we will also need the following theorem, that a function must be continuous wherever the derivative is defined.

Because $f'(c)$ is defined, the limit defining it is

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

We know that $\displaystyle \lim_{x \to c} x-c = c-c = 0$ using the Limit of a Linear Function. Because

\begin{equation*} f(x) - f(c) = \frac{f(x)-f(c)}{x-c} \cdot (x-c), \end{equation*}

we can compute the limit

\begin{equation*} \lim_{x \to c} f(x)-f(c) = \lim_{x \to c} \left[ \frac{f(x)-f(c)}{x-c} \cdot (x-c) \right] = f'(c) \cdot 0 = 0. \end{equation*}

The value $f(c)$ is a constant, so the Limit of a Constant Rule implies

\begin{equation*} \lim_{x \to c} f(c) = f(c). \end{equation*}

Since $f(x)=(f(x)-f(c)) + f(c)\text{,}$ the Limit of a Sum Rule implies

\begin{equation*} \lim_{x \to c} f(x) = \lim_{x \to c} [ f(x)-f(c)+f(c)] = 0 + f(c) = f(c). \end{equation*}

Therefore, $f$ is continuous at $c\text{.}$

### Subsection8.5.2Proofs of Differentiation Rules

#### Subsubsection8.5.2.1Proof of Constant Multiple Rule

By hypothesis, $\displaystyle \frac{d}{dx}[f(x)] = f'(x)\text{.}$ This means that $f'(x)$ is defined by its limit

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

The rule is interested in finding the rate of change of a new function $k \cdot f(x)\text{.}$ We will use that function to compute the derivative.

\begin{align*} \frac{d}{dx}[k \cdot f(x)] &= \lim_{h \to 0} \frac{k \cdot f(x+h) - k \cdot f(x)}{h} \\ &= \lim_{h \to 0} \frac{k (f(x+h)-f(x))}{h}. \end{align*}

Now notice that the formula is a product of the constant $k$ and the average rate of change of $f\text{.}$ Because we already know that the limit of the average rate of change of $f$ is equal to $f'(x)\text{,}$ we can use the Limit Rule for a Constant Multiple.

\begin{align*} \frac{d}{dx}[k \cdot f(x)] &= \lim_{h \to 0} \frac{k (f(x+h)-f(x))}{h} \\ &= k \cdot f'(x). \end{align*}

#### Subsubsection8.5.2.2Proof of Reciprocal Rule

By hypothesis, $\displaystyle \frac{d}{dx}[g(x)] = g'(x)\text{.}$ This means that $g'(x)$ is defined by its limit

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

The rule is interested in finding the rate of change of a new function $1/g(x)\text{.}$ We will use that function to compute the derivative using the definition, which will require finding a common denominator.

\begin{align*} \frac{d}{dx}[\frac{1}{g(x)}] &= \lim_{h \to 0} \frac{\frac{1}{g(x+h)} - \frac{1}{g(x)}}{h} \\ &= \lim_{h \to 0} \frac{\frac{g(x)}{g(x)g(x+h)} - \frac{g(x+h)}{g(x)g(x+h)}}{h} \\ &= \lim_{h \to 0} \frac{g(x)-g(x+h)}{g(x)g(x+h)}\cdot \frac{1}{h} \\ &= \lim_{h \to 0} \frac{-1}{g(x)g(x+h)}\cdot \frac{g(x+h)-g(x)}{h} \end{align*}

Since the limit involves $h \to 0\text{,}$ $g(x)$ is a constant and $g(x+h) \to g(x)$ (by continuity) so that

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

Using the Limit Rule of a Product, we have

\begin{align*} \frac{d}{dx}[\frac{1}{g(x)}] &= \lim_{h \to 0} \frac{-1}{g(x)g(x+h)}\cdot \frac{g(x+h)-g(x)}{h} \\ &= \frac{-1}{(g(x))^2} \cdot g'(x) = \frac{-g'(x)}{(g(x))^2}. \end{align*}

#### Subsubsection8.5.2.3Proof of Sum Rule

By hypothesis, $\displaystyle \frac{d}{dx}[f(x)] = f'(x)$ and $\displaystyle \frac{d}{dx}[g(x)] = g'(x)\text{.}$ This means that

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

The sum rule is interested in finding the rate of change of a new function $f(x)+g(x)\text{.}$ We will use that function to compute the derivative using the definition:

\begin{align*} \frac{d}{dx}[f(x)+g(x)] &= \lim_{h \to 0} \frac{[f(x+h)+g(x+h)] - [f(x)+g(x)]}{h} \\ &= \lim_{h \to 0} \frac{f(x+h)+g(x+h)-f(x)-g(x)}{h} \\ &= \lim_{h \to 0} \frac{f(x+h)-f(x)+g(x+h)-g(x)}{h} \\ &= \lim_{h \to 0} \left[\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}\right] \\ &= f'(x)+g'(x), \end{align*}

using the Limit Rule of a Sum.

#### Subsubsection8.5.2.4Proof of Product Rule

By hypothesis, $\displaystyle \frac{d}{dx}[f(x)] = f'(x)$ and $\displaystyle \frac{d}{dx}[g(x)] = g'(x)\text{.}$ This means that

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

The product rule is interested in finding the rate of change of a new function $f(x)g(x)\text{.}$ In the course of this calculation, we will also need the following substitutions,

\begin{gather*} f(x+h) = f(x+h)-f(x) + f(x) = \Delta f + f(x), \\ g(x+h) = g(x+h)-g(x) + g(x) = \Delta g + g(x). \end{gather*}

When $f'(x)$ and $g'(x)$ both exist, $f$ and $g$ are both continuous so that

\begin{gather*} \lim_{h \to 0} \Delta f = \lim_{h \to 0} f(x+h)-f(x) = 0, \\ \lim_{h \to 0} \Delta g = \lim_{h \to 0} g(x+h)-g(x) = 0. \end{gather*}

The derivative in question is defined by

\begin{align*} \frac{d}{dx}[f(x)g(x)] &= \lim_{h \to 0} \frac{[f(x+h)g(x+h)] - [f(x)g(x)]}{h} \\ &= \lim_{h \to 0} \frac{(\Delta f + f(x))(\Delta g + g(x)) - f(x)g(x)}{h} \\ &= \lim_{h \to 0} \frac{\Delta f \Delta g + \Delta f g(x) + f(x) \Delta g + f(x)g(x) - f(x)g(x)}{h} \\ &= \lim_{h \to 0} \frac{\Delta f \Delta g + \Delta f g(x) + f(x) \Delta g}{h} \\ &= \lim_{h \to 0} \left[\frac{\Delta f \Delta g}{h} + \frac{\Delta f g(x)}{h} + \frac{f(x) \Delta g}{h} \right]. \\ &= \lim_{h \to 0} \left[\Delta f \cdot \frac{\Delta g}{h} + \frac{\Delta f}{h} \cdot g(x) + f(x) \cdot \frac{\Delta g}{h} \right]. \\ &= 0 \cdot g'(x) + f'(x) \cdot g(x) + f(x) \cdot g'(x) \\ &= f'(x) \cdot g(x) + f(x) \cdot g'(x), \end{align*}

using the Limit Rule of a Sum and the Limit Rule of a Product.

### Subsection8.5.3Using the Derivative Rules

A traditional development of calculus begins applying these rules to find formulas to derivatives. We will instead begin applying the rules by interpreting some specific applied applications involving rates of change.

###### Example8.5.4

A tank is being filled with water two supply hoses. If the first hose is pumping water at a rate of 20 gal/min and the second hose is pumping water at a rate of 30 gal/min, what is the total rate of change for the tank?

Solution

We know the intuitive solution to the problem is 50 gal/min. This is actually a consequence of the sum rule of derivatives.

We can think of the water in the tank as having two components: $W_1\text{,}$ the volume of water (gal) that was pumped by hose 1, and $W_2\text{,}$ the volume of water (gal) that was pumped by hose 2. These two variables are functions of time $t$ (min), and the rates of water flowing from the hoses correspond to derivatives:

\begin{equation*} \frac{dW_1}{dt} = 20, \quad \frac{dW_2}{dt} = 30. \end{equation*}

The total volume of water in the tank at a given time $t$ is the sum $W(t)=W_1(t) + W_2(t)\text{.}$ So by the sum rule of derivatives,

\begin{equation*} \frac{dW}{dt} = \frac{dW_1}{dt} + \frac{dW_2}{dt} = 20+30 = 50. \end{equation*}

The sum rule for derivatives feels very intuitive. If a quantity is the sum of parts, then the total rate of change for the quantity is the sum of the rates of change for each of the parts. The product rule is less intuitive because we don't get to multiply rates of change when a quantity is a product. To illustrate this example, we focus on a geometric example on the area of a rectangle when the lengths of the sides are changing.

###### Example8.5.5

A city is in the shape of a rectangle with sides aligned with North-South and East-West lines. Suppose that the city is currently 5 miles east-to-west and 3 miles north-to-south and plans to expand to a size 8 miles east-to-west by 5 miles north-to-south over the next 10 years. What is the average rate of change of the total area in the city over the 10 years? If the borders were to move at a constant rate over those 10 years, what is the instantaneous rate of change of the total area of the city at the beginning and at the end of the 10 years?

Solution

We start with the question of average rate of change of total area, which we can solve intuitively. The city originally has a total area of $5 \times 3 = 15 \: \hbox{mi}^2\text{.}$ After 10 years, the city has a total area of $8 \times 5 = 40 \: \hbox{mi}^2\text{.}$ So the change in area is $25 \: \hbox{mi}^2$ over 10 years so the average rate of change of area is $\frac{25}{10} = 2.5\: \hbox{mi}^2/\hbox{yr}\text{.}$

To connect our intuition with functions and to prepare for the next calculations, let us introduce some variables. The state of the city can be characterized by four variables: the time $t$ (yr) at which the state is observed, the distance east-to-west, which we'll call the width $W$ (mi), the distance north-to-south, which we'll call the height $H$ (mi), and the enclosed total area $A$ ($\hbox{mi}^2$). We think of $W\text{,}$ $H$ and $A$ as being functions of time $t$ with the area being equal to the product of $W$ and $H\text{:}$

\begin{equation*} A(t) = W(t) \cdot H(t). \end{equation*}

Then the average rate of change for $A$ on the interval $t \in [0,10]$ is given by

\begin{equation*} \left. \frac{\Delta A}{\Delta t} \right|_{[0,10]} = \frac{A(10)-A(0)}{10-0} = \frac{W(10)H(10) - W(0)H(0)}{10} = \frac{8(5)-5(3)}{10} = 2.5. \end{equation*}

To find the instantaneous rates of change, we need to know how fast the width and height measurements are changing in time. Because the problem stated that these changed at a constant rate, we can use the average rates of change to compute the instantaneous rates:

\begin{gather*} \displaystyle \frac{dW}{dt} = \left. \frac{\Delta W}{\Delta t}\right|_{[0,10]} = \frac{W(10)-W(0)}{10-0} = \frac{8-5}{10} = 0.3, \\ \displaystyle \frac{dH}{dt} = \left. \frac{\Delta H}{\Delta t}\right|_{[0,10]} = \frac{H(10)-H(0)}{10-0} = \frac{5-3}{10} = 0.2. \end{gather*}

Since the area $A$ is the product of $W$ and $H\text{,}$ the product rule for derivatives will provide the instantaneous rate of change for area:

\begin{equation*} \frac{dA}{dt} = \frac{d}{dt}[W \cdot H] = \frac{dW}{dt} \cdot H + W \cdot \frac{dH}{dt}. \end{equation*}

When $t=0$ we have $W(0)=5$ and $H(0)=3$ so that

\begin{align*} \left. \frac{dA}{dt} \right|_{0} &= \left.\frac{dW}{dt}\right|_{0} \cdot H(0) + W(0) \cdot \left.\frac{dH}{dt}\right|_{0}\\ &= 0.3(3) + 5(0.2) = 1.9. \end{align*}

That is, at the beginning, the city is expanding at a rate of $1.9 \: \hbox{mi}^2/\hbox{yr}\text{.}$ After 10 years, $t=10\text{,}$ we have $W(10)=8$ and $H(10)=5$ so that

\begin{align*} \left. \frac{dA}{dt} \right|_{10} &= \left.\frac{dW}{dt}\right|_{10} \cdot H(10) + W(10) \cdot \left.\frac{dH}{dt}\right|_{10}\\ &= 0.3(5) + 8(0.2) = 3.1. \end{align*}

At the end of the 10 years, the city is expanding at a rate of $3.1 \: \hbox{mi}^2/\hbox{yr}\text{.}$

The picture of expanding area helps provide some intuition for why the product rule is the appropriate technique. If we consider the city after 6 months ($t=0.5$), both the width and the height have changed by a small amount, as shown in the figure below. The total change in area has two primary contributions, corresponding to long, skinny rectangles with areas $W(0) \cdot \Delta H$ and $\Delta W \cdot H(0)\text{,}$ and a very small rectangle with area $\Delta W \cdot \Delta H\text{.}$ The product rule corresponds to the rate of change coming from the two primary contributions while the small rectangle leads to a term that has a limit of zero in the calculation of the derivative.

Quotients often appear when working with densities, concentrations, or other ratios.

###### Example8.5.6

A salt-water solution is being formulated. At a particular instant, the solution consists of 10 L of water with 5 kg of salt. At that instant, water is being added at a rate of 0.5 L/s while salt is being added at a rate of 0.2 kg/s. What is the instantaneous rate of change of the concentration?

Solution

We start by identifying the variables that define the state of our system. The variables include the time $t\text{,}$ measured in seconds (s), the total volume of water $V\text{,}$ measured in liters (L), the total amount of salt in the water $S\text{,}$ measured in kilograms (kg), and the concentration of salt water $C\text{,}$ measured in kilograms per liter (kg/L). The variables $V\text{,}$ $S$ and $C$ are functions of time $t$ with an equation relating them by

The instantaneous rate of change is computed using the quotient rule for derivatives,

\begin{equation*} \frac{dC}{dt} = \frac{ V \frac{dS}{dt} - S \frac{dV}{dt}}{V^2}. \end{equation*}

The values at the instant in question are given by

\begin{gather*} V = 10, \quad \frac{dV}{dt} = 0.5, \\ S = 5, \quad \frac{dS}{dt} = 0.2. \end{gather*}

Using these values in the quotient rule for derivatives, we have

\begin{equation*} \frac{dC}{dt} = \frac{ 10(0.2) - 5(0.5) }{10^2} = \frac{2-2.5}{100} = -0.005. \end{equation*}

That is, the concentration is changing at a rate of -0.005 kg salt per liter water per second. Alternatively, we could say that the concentration is decreasing at a rate of 0.005 kg/L/s.

### Subsection8.5.4Derivative Building Blocks

In order to apply the differentiation rules for formulas, we need to have some elementary rules to get started. Just as with the limit rules, we begin with the basics. The justification of the derivatives of the elementary derivatives must be based on the definition of the derivative. The derivatives of more complex functions can then be justified using the derivative rules.

Let $f(x)=k$ be the constant function. Since $f(x+h)=k\text{,}$ we have

\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac{k-k}{h} = \lim_{h \to 0} 0 \\ &= 0, \end{align*}

where the last step used the limit rule for a constant.

Let $f(x)=x$ be the identity function. Since $f(x+h)=x+h\text{,}$ we have

\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac{x+h-x}{h} = \lim_{h \to 0} \frac{h}{h} \\ &= \lim_{h \to 0} 1 = 1. \end{align*}

Although the derivatives of other functions can be found using the limit definition of the derivative, these two elementary derivatives along with the the rules of differentiation allow us to find other derivatives.

###### Example8.5.9
Find $f'(x)$ for $f(x) = 3x-5\text{,}$ citing the differentiation rules you used.
Solution

We will use the differentiation operator $\frac{d}{dx}\text{.}$ Recall that this operator behaves something like a function, but with a function as its input and the derivative of that function as an output. The rules of differentiation allow us to find a derivative by breaking the function down into its elementary components.

In this example, $f(x)$ is a sum of $3x$ and -5. The subformula $3x$ is a constant multiple (3) of the identity $x$ and the number -5 is a constant. Once you recognize these facts, you apply the rules.

\begin{align*} f'(x) &= \frac{d}{dx}[3x-5] = \frac{d}{dx}[3x+-5] \\ &= \frac{d}{dx}[3x] + \frac{d}{dx}[-5] \quad \hbox{Sum Rule} \\ &= 3 \frac{d}{dx}[x] + \frac{d}{dx}[-5] \quad \hbox{Constant Multiple Rule} \\ &= 3 \cdot 1 + \frac{d}{dx}[-5] \quad \hbox{Derivative of Identity} \\ &= 3 \cdot 1 + 0 \quad \hbox{Derivative of Constant} \\ &= 3 \end{align*}

Thus, $f'(x) = 3\text{.}$

The rules of derivatives can also allow us to create new differentiation rules. The following theorem uses the same steps as the previous example but with arbitrary constants.

\begin{align*} \frac{d}{dx}[ax+b] &= \frac{d}{dx}[ax] + \frac{d}{dx}[b] \quad \hbox{Sum Rule} \\ &= a \frac{d}{dx}[x] + \frac{d}{dx}[b] \quad \hbox{Constant Multiple Rule} \\ &= a \cdot 1 + \frac{d}{dx}[b] \quad \hbox{Derivative of Identity} \\ &= a \cdot 1 + 0 \quad \hbox{Derivative of Constant} \\ &= a \end{align*}

Integer powers correspond to repeated multiplication, so the product rule of differentiation will lead to a rule for the derivative of a power. The following examples lead to a natural pattern called the power rule for derivatives.

###### Example8.5.11
Use the product rule of derivatives to show that
\begin{gather*} \frac{d}{dx}[x^2] = 2x ,\\ \frac{d}{dx}[x^3] = 3x^2, \\ \frac{d}{dx}[x^4] = 4x^3. \end{gather*}
Solution

The work will involve rewriting the powers as products,

\begin{equation*} x^2 = x \cdot x, \quad x^3 = x \cdot x^2, \quad x^4 = x \cdot x^3. \end{equation*}

\begin{equation*} \frac{d}{dx}[x] = 1. \end{equation*}

It is useful to remember the product rule using dependent variables, say $u=f(x)$ and $v=g(x)\text{,}$ as

\begin{equation*} \frac{d}{dx}[u \cdot v] = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \end{equation*}

because this will guide our use of the differentiation operator.

With that setup, we can compute the derivatives using differentiation rules.

\begin{align*} \frac{d}{dx}[x^2] &= \frac{d}{dx}[x \cdot x]\\ &= \frac{d}{dx}[x] \cdot x + x \cdot \frac{d}{dx}[x]\\ &= 1 \cdot x + x \cdot 1\\ &= 2x \end{align*}

For the next calculation, we will use both of our previously found derivatives.

\begin{align*} \frac{d}{dx}[x^3] &= \frac{d}{dx}[x \cdot x^2]\\ &= \frac{d}{dx}[x] \cdot x^2 + x \cdot \frac{d}{dx}[x^2]\\ &= 1 \cdot x^2 + x \cdot (2x)\\ &= 3x^2 \end{align*}

The pattern should be apparent for the next derivative.

\begin{align*} \frac{d}{dx}[x^4] &= \frac{d}{dx}[x \cdot x^3]\\ &= \frac{d}{dx}[x] \cdot x^3 + x \cdot \frac{d}{dx}[x^3]\\ &= 1 \cdot x^3 + x \cdot (3x^2)\\ &= 4x^3 \end{align*}

We can continue to find more derivatives using these results.

###### Example8.5.12
Find $\displaystyle \frac{d}{dx}[\frac{1}{x^3}]\text{.}$
Solution
\begin{equation*} \frac{d}{dx}[\frac{1}{u}] = \frac{-\frac{du}{dx}}{u^2}. \end{equation*}
\begin{align*} \frac{d}{dx}[\frac{1}{x^3}] &= \frac{-\frac{d}{dx}[x^3]}{(x^3)^2} \\ &= \frac{-(3x^2)}{x^3 \cdot x^3} = \frac{-3x^2}{x^6} \\ &= \frac{-3}{x^4}. \end{align*}
###### Example8.5.13
Find $\displaystyle \frac{d}{dx}[ 5x^2-8x+3 ]\text{.}$
Solution
$f(x)=5x^2-8x+3\text{.}$
\begin{align*} f'(x) &= \frac{d}{dx}[5x^2-8x+3] \\ & = \frac{d}{dx}[5x^2 + (-8x+3)] \\ & = \frac{d}{dx}[5x^2] + \frac{d}{dx}[-8x+3] \end{align*}
$x^2$
\begin{align*} f'(x) & = 5 \frac{d}{dx}[x^2] + \frac{d}{dx}[-8x+3] \\ & = 5(2x) + -8\\ & = 10x-8. \end{align*}
$\displaystyle \frac{d}{dx}[ 5x^2-8x+3 ] = 10x-8\text{.}$