3.3 Differentiation Rules

Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather challenging process. For example, previously we found that $\frac{d}{d x} (x) = \frac{1}{2 x}$ by using a process that involved multiplying an expression by a conjugate prior to evaluating a limit. The process that we could use to evaluate $\frac{d}{d x} (3 x)$ using the definition, while similar, is more complicated. In this section, we develop rules for finding derivatives that allow us to bypass this process. We begin with the basics.

The Basic Rules

The functions $f (x) = c$ and $g (x) = x^{n}$ where $n$ is a positive integer are the building blocks from which all polynomials and rational functions are constructed. To find derivatives of polynomials and rational functions efficiently without resorting to the limit definition of the derivative, we must first develop formulas for differentiating these basic functions.

The Constant Rule

We first apply the limit definition of the derivative to find the derivative of the constant function, $f (x) = c .$ For this function, both $f (x) = c$ and $f (x + h) = c,$ so we obtain the following result:

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} \frac{0}{h} = lim_{h \to 0} 0 = 0.

The rule for differentiating constant functions is called the constant rule. It states that the derivative of a constant function is zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function is $0.$ We restate this rule in the following theorem.

The Power Rule

We have shown that

\frac{d}{d x} (x^{2}) = 2 x and \frac{d}{d x} (x^{1 / 2}) = \frac{1}{2} x^{- 1 / 2} .

At this point, you might see a pattern beginning to develop for derivatives of the form $\frac{d}{d x} (x^{n}) .$ We continue our examination of derivative formulas by differentiating power functions of the form $f (x) = x^{n}$ where $n$ is a positive integer. We develop formulas for derivatives of this type of function in stages, beginning with positive integer powers. Before stating and proving the general rule for derivatives of functions of this form, we take a look at a specific case, As we go through this derivation, note that the technique used in this case is essentially the same as the technique used to prove the general case.

As we shall see, the procedure for finding the derivative of the general form $f (x) = x^{n}$ is very similar. Although it is often unwise to draw general conclusions from specific examples, we note that when we differentiate $f (x) = x^{3},$ the power on $x$ becomes the coefficient of $x^{2}$ in the derivative and the power on in the derivative decreases by 1. The following theorem states that the holds for all positive integer powers of We will eventually extend this result to negative integer powers. Later, we will see that this rule may also be extended first to rational powers of and then to arbitrary powers of Be aware, however, that this rule does not apply to functions in which a constant is raised to a variable power, such as

Proof

For $f (x) = x^{n}$ where $n$ is a positive integer, we have

f^{'} (x) = h \to 0 lim \frac{( x + h ) ^{n} - x ^{n}}{h} .

Since (x + h)^{n} = x^{n} + n x^{n - 1} h + (n 2) x^{n - 2} h^{2} + (n 3) x^{n - 3} h^{3} + \dots + n x h^{n - 1} + h^{n},

we see that

(x + h)^{n} - x^{n} = n x^{n - 1} h + (n 2) x^{n - 2} h^{2} + (n 3) x^{n - 3} h^{3} + \dots + n x h^{n - 1} + h^{n} .

Next, divide both sides by h:

\frac{( x + h ) ^{n} - x ^{n}}{h} = \frac{n x ^{n - 1} h + ( n 2 ) x ^{n - 2} h ^{2} + ( n 3 ) x ^{n - 3} h ^{3} + \dots + n x h ^{n - 1} + h ^{n}}{h} .

Thus,

\frac{( x + h ) ^{n} - x ^{n}}{h} = n x^{n - 1} + (n 2) x^{n - 2} h + (n 3) x^{n - 3} h^{2} + \dots + n x h^{n - 2} + h^{n - 1} .

Finally,

f^{'} (x) = lim_{h \to 0} (n x^{n - 1} + (n 2) x^{n - 2} h + (n 3) x^{n - 3} h^{2} + \dots + n x h^{n - 2} + h^{n - 1}) = n x^{n - 1} .

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The Sum, Difference, and Constant Multiple Rules

We find our next differentiation rules by looking at derivatives of sums, differences, and constant multiples of functions. Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract, or multiply by a constant. These rules are summarized in the following theorem.

Proof

We provide only the proof of the sum rule here. The rest follow in a similar manner.

For differentiable functions $f (x)$ and $g (x),$ we set $j (x) = f (x) + g (x) .$ Using the limit definition of the derivative we have

j^{'} (x) = h \to 0 lim \frac{j ( x + h ) - j ( x )}{h} .

By substituting $j (x + h) = f (x + h) + g (x + h)$ and $j (x) = f (x) + g (x),$ we obtain

j^{'} (x) = h \to 0 lim \frac{( f ( x + h ) + g ( x + h )) - ( f ( x ) + g ( x ))}{h} .

Rearranging and regrouping the terms, we have

j^{'} (x) = h \to 0 lim (\frac{f ( x + h ) - f ( x )}{h} + \frac{g ( x + h ) - g ( x )}{h}) .

We now apply the sum law for limits and the definition of the derivative to obtain

j^{'} (x) = h \to 0 lim (\frac{f ( x + h ) - f ( x )}{h}) + h \to 0 lim (\frac{g ( x + h ) - g ( x )}{h}) = f^{'} (x) + g^{'} (x) .

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The Product Rule

Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern. To see why we cannot use this pattern, consider the function $f (x) = x^{2},$ whose derivative is $f^{'} (x) = 2 x$ and not $\frac{d}{d x} (x) \cdot \frac{d}{d x} (x) = 1 \cdot 1 = 1.$

Proof

We begin by assuming that $f (x)$ and $g (x)$ are differentiable functions. At a key point in this proof we need to use the fact that, since $g (x)$ is differentiable, it is also continuous. In particular, we use the fact that since $g (x)$ is continuous, $lim_{h \to 0} g (x + h) = g (x) .$

By applying the limit definition of the derivative to $j (x) = f (x) g (x),$ we obtain

j^{'} (x) = h \to 0 lim \frac{f ( x + h ) g ( x + h ) - f ( x ) g ( x )}{h} .

By adding and subtracting $f (x) g (x + h)$ in the numerator, we have

j^{'} (x) = h \to 0 lim \frac{f ( x + h ) g ( x + h ) - f ( x ) g ( x + h ) + f ( x ) g ( x + h ) - f ( x ) g ( x )}{h} .

After breaking apart this quotient and applying the sum law for limits, the derivative becomes

j^{'} (x) = h \to 0 lim (\frac{f ( x + h ) g ( x + h ) - f ( x ) g ( x + h )}{h}) + h \to 0 lim (\frac{f ( x ) g ( x + h ) - f ( x ) g ( x )}{h}) .

Rearranging, we obtain

j^{'} (x) = h \to 0 lim (\frac{f ( x + h ) - f ( x )}{h} \cdot g (x + h)) + h \to 0 lim (\frac{g ( x + h ) - g ( x )}{h} \cdot f (x)) .

By using the continuity of $g (x),$ the definition of the derivatives of $f (x)$ and $g (x),$ and applying the limit laws, we arrive at the product rule,

j^{'} (x) = f^{'} (x) g (x) + g^{'} (x) f (x) .

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The Quotient Rule

Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that

\frac{d}{d x} (x^{2}) = 2 x, not \frac{\frac{d}{d x} ( x ^{3} )}{\frac{d}{d x} ( x )} = \frac{3 x ^{2}}{1} = 3 x^{2} .

The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form $x^{k}$ where $k$ is a negative integer.

Proof

If $k$ is a negative integer, we may set $n = - k,$ so that n is a positive integer with $k = - n .$ Since for each positive integer $n, x^{- n} = \frac{1}{x ^{n}},$ we may now apply the quotient rule by setting and In this case, and Thus,

\frac{d}{d x} (x^{- n}) = \frac{0 ( x ^{n} ) - 1 ( n x ^{n - 1} )}{( x ^{n} ) ^{2}} .

Simplifying, we see that

\frac{d}{d x} (x^{- n}) = \frac{- n x ^{n - 1}}{x ^{2 n}} = - n x^{(n - 1) - 2 n} = - n x^{- n - 1} .

Finally, observe that since $k = - n,$ by substituting we have

\frac{d}{d x} (x^{k}) = k x^{k - 1} .

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Combining Differentiation Rules

As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function.

Student ProjectFormula One Grandstands

Formula One car races can be very exciting to watch and attract a lot of spectators. Formula One track designers have to ensure sufficient grandstand space is available around the track to accommodate these viewers. However, car racing can be dangerous, and safety considerations are paramount. The grandstands must be placed where spectators will not be in danger should a driver lose control of a car (Figure 3.20).

Figure 3.20The grandstand next to a straightaway of the Circuit de Barcelona-Catalunya race track, located where the spectators are not in danger.

Safety is especially a concern on turns. If a driver does not slow down enough before entering the turn, the car may slide off the racetrack. Normally, this just results in a wider turn, which slows the driver down. But if the driver loses control completely, the car may fly off the track entirely, on a path tangent to the curve of the racetrack.

Suppose you are designing a new Formula One track. One section of the track can be modeled by the function $f (x) = x^{3} + 3 x^{2} + x$ (Figure 3.21). The current plan calls for grandstands to be built along the first straightaway and around a portion of the first curve. The plans call for the front corner of the grandstand to be located at the point We want to determine whether this location puts the spectators in danger if a driver loses control of the car.

Key Concepts

The derivative of a constant function is zero.
The derivative of a power function is a function in which the power on $x$ becomes the coefficient of the term and the power on $x$ in the derivative decreases by 1.
The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative.
The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g.
The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g.
The derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.
The derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function.
We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resorting to the definition of the derivative. These formulas can be used singly or in combination with each other.

The Basic Rules

The Constant Rule

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} \frac{0}{h} = lim_{h \to 0} 0 = 0.

The Power Rule

We have shown that

\frac{d}{d x} (x^{2}) = 2 x and \frac{d}{d x} (x^{1 / 2}) = \frac{1}{2} x^{- 1 / 2} .

Proof

For $f (x) = x^{n}$ where $n$ is a positive integer, we have

f^{'} (x) = h \to 0 lim \frac{( x + h ) ^{n} - x ^{n}}{h} .

Since (x + h)^{n} = x^{n} + n x^{n - 1} h + (n 2) x^{n - 2} h^{2} + (n 3) x^{n - 3} h^{3} + \dots + n x h^{n - 1} + h^{n},

we see that

(x + h)^{n} - x^{n} = n x^{n - 1} h + (n 2) x^{n - 2} h^{2} + (n 3) x^{n - 3} h^{3} + \dots + n x h^{n - 1} + h^{n} .

Next, divide both sides by h:

\frac{( x + h ) ^{n} - x ^{n}}{h} = \frac{n x ^{n - 1} h + ( n 2 ) x ^{n - 2} h ^{2} + ( n 3 ) x ^{n - 3} h ^{3} + \dots + n x h ^{n - 1} + h ^{n}}{h} .

Thus,

\frac{( x + h ) ^{n} - x ^{n}}{h} = n x^{n - 1} + (n 2) x^{n - 2} h + (n 3) x^{n - 3} h^{2} + \dots + n x h^{n - 2} + h^{n - 1} .

Finally,

f^{'} (x) = lim_{h \to 0} (n x^{n - 1} + (n 2) x^{n - 2} h + (n 3) x^{n - 3} h^{2} + \dots + n x h^{n - 2} + h^{n - 1}) = n x^{n - 1} .

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The Sum, Difference, and Constant Multiple Rules

Proof

We provide only the proof of the sum rule here. The rest follow in a similar manner.

For differentiable functions $f (x)$ and $g (x),$ we set $j (x) = f (x) + g (x) .$ Using the limit definition of the derivative we have

j^{'} (x) = h \to 0 lim \frac{j ( x + h ) - j ( x )}{h} .

By substituting $j (x + h) = f (x + h) + g (x + h)$ and $j (x) = f (x) + g (x),$ we obtain

j^{'} (x) = h \to 0 lim \frac{( f ( x + h ) + g ( x + h )) - ( f ( x ) + g ( x ))}{h} .

Rearranging and regrouping the terms, we have

j^{'} (x) = h \to 0 lim (\frac{f ( x + h ) - f ( x )}{h} + \frac{g ( x + h ) - g ( x )}{h}) .

We now apply the sum law for limits and the definition of the derivative to obtain

j^{'} (x) = h \to 0 lim (\frac{f ( x + h ) - f ( x )}{h}) + h \to 0 lim (\frac{g ( x + h ) - g ( x )}{h}) = f^{'} (x) + g^{'} (x) .

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The Product Rule

Proof

By applying the limit definition of the derivative to $j (x) = f (x) g (x),$ we obtain

j^{'} (x) = h \to 0 lim \frac{f ( x + h ) g ( x + h ) - f ( x ) g ( x )}{h} .

By adding and subtracting $f (x) g (x + h)$ in the numerator, we have

j^{'} (x) = h \to 0 lim \frac{f ( x + h ) g ( x + h ) - f ( x ) g ( x + h ) + f ( x ) g ( x + h ) - f ( x ) g ( x )}{h} .

After breaking apart this quotient and applying the sum law for limits, the derivative becomes

j^{'} (x) = h \to 0 lim (\frac{f ( x + h ) g ( x + h ) - f ( x ) g ( x + h )}{h}) + h \to 0 lim (\frac{f ( x ) g ( x + h ) - f ( x ) g ( x )}{h}) .

Rearranging, we obtain

j^{'} (x) = h \to 0 lim (\frac{f ( x + h ) - f ( x )}{h} \cdot g (x + h)) + h \to 0 lim (\frac{g ( x + h ) - g ( x )}{h} \cdot f (x)) .

By using the continuity of $g (x),$ the definition of the derivatives of $f (x)$ and $g (x),$ and applying the limit laws, we arrive at the product rule,

j^{'} (x) = f^{'} (x) g (x) + g^{'} (x) f (x) .

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The Quotient Rule

\frac{d}{d x} (x^{2}) = 2 x, not \frac{\frac{d}{d x} ( x ^{3} )}{\frac{d}{d x} ( x )} = \frac{3 x ^{2}}{1} = 3 x^{2} .

The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form $x^{k}$ where $k$ is a negative integer.

Proof

\frac{d}{d x} (x^{- n}) = \frac{0 ( x ^{n} ) - 1 ( n x ^{n - 1} )}{( x ^{n} ) ^{2}} .

Simplifying, we see that

\frac{d}{d x} (x^{- n}) = \frac{- n x ^{n - 1}}{x ^{2 n}} = - n x^{(n - 1) - 2 n} = - n x^{- n - 1} .

Finally, observe that since $k = - n,$ by substituting we have

\frac{d}{d x} (x^{k}) = k x^{k - 1} .

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Combining Differentiation Rules

Student ProjectFormula One Grandstands

Figure 3.20The grandstand next to a straightaway of the Circuit de Barcelona-Catalunya race track, located where the spectators are not in danger.

Key Concepts

The derivative of a constant function is zero.
The derivative of a power function is a function in which the power on $x$ becomes the coefficient of the term and the power on $x$ in the derivative decreases by 1.
The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative.
The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g.
The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g.
The derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.
The derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function.
We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resorting to the definition of the derivative. These formulas can be used singly or in combination with each other.