Techniques of differentiation

An aerial view of a residential neighborhood with solar panels on several rooftops — Photo by Kevin Martin Jose on Unsplash.

So far, we've learned how to differentiate familiar power functions and polynomials one term at a time. But many functions are not presented as a simple sum of terms: different pieces can be multiplied or divided, and these pieces may be nested in layers. How can we extend what we already know to functions built from many smaller pieces?

In this lesson, we'll learn to follow small changes through the structure of a function, handling one piece at a time and then assembling the results. This perspective turns our collection of derivative formulas into a more flexible problem-solving system, one that will support nearly every differentiation technique that follows.

By the end of this lesson, you should be able to:

Differentiate composite functions by following changes through one layer at a time.
Differentiate products and quotients, and explain how each part contributes to the result.
Choose and combine differentiation rules to analyze functions built from smaller pieces.

Suppose that a rapidly growing metropolitan area is trying to anticipate how much electricity its residents will use. Regional planners decide to model this situation with the equation $Q=H\cdot E\text{,}$ where $$ Q $$ is the metropolitan area's annual residential electricity use, $$ H $$ is the number of households in the metropolitan area, and $$ E $$ is the average annual electricity use per household.

Since $$ Q $$ depends on both $$ H $$ and $$ E $$ , a change in either quantity will affect the overall electricity use. For example:

If the number of households increases while the average electricity use per household stays fixed, then total use .
If average electricity use per household decreases while the number of households stays fixed, then total use .

Those cases are straightforward because only one quantity changes at a time, but what happens when both quantities change?

To get an idea of how to tackle this problem, let's take a concrete example. We'll use the following units:

$$ H $$ will be measured in thousands of households; that is, $$ H=40 $$ corresponds to $$ 40{,}000 $$ households.
$$ E $$ will be measured in megawatt-hours (MWh) of annual electricity use per household. (A megawatt-hour is the amount of energy produced if 1 million watts of power are generated continuously over the course of one hour.)
$$ Q $$ will therefore be measured in gigawatt-hours (GWh), since a gigawatt is a thousand megawatts.

So if we have $$ 40 $$ thousand households, each using an average of $$ 12 $$ MWh annually, then the metropolitan area's annual residential electricity use is $$ Q= $$ GWh.

Now, suppose the metropolitan area gains $$ 2 $$ thousand households, while average annual use increases by $$ 0.5 $$ MWh per household. We could substitute these new values into $Q=H\cdot E$ , but that calculation would only tell us the total change; instead, let's look specifically at how the two changing quantities contribute to it.

A useful problem-solving strategy in mathematics is to try out multiple representations of a problem, as sometimes finding just the representation can give new insight. One useful way to visualize multiplication is by thinking of it as area, so we'll imagine a rectangle with sides $$ H $$ and $$ E $$ , which the product $Q=H\cdot E$ can be thought of as the area of that rectangle. Then the pieces added to the rectangle when we change $$ H $$ and $$ E $$ will let us see how $$ Q $$ changes.

Loading the graph…If it does not appear, check the Desmos API configuration.

Move the slider to increase the number of households from

\( 40 \)

thousand to

\( 42 \)

thousand and the average annual use from

\( 12 \)

\( 12.5 \)

MWh per household. The rectangle is not drawn to scale.

The original rectangle represents $$ 480 $$ GWh of annual use. Changing both sides adds three pieces. Use the labeled dimensions to find the electricity use represented by each piece:

The right strip with dimensions $E\cdot\Delta H$ has an area of .
The top strip with dimensions $H\cdot\Delta E$ has an area of .
The corner with dimensions $\Delta H\cdot\Delta E$ has an area of .

Together, these pieces show that the exact increase in annual electricity use is GWh.

From this numerical example, we can see a general pattern: if $$ H $$ changes by $\Delta H$ and $$ E $$ changes by $\Delta E$ , then

$\begin{aligned}\Delta Q &=(H+\Delta H)(E+\Delta E)-HE\\ &=\mathhlblue{E\cdot\Delta H}+\mathhlgreen{H\cdot\Delta E}+\mathhlviolet{\Delta H\cdot\Delta E}\text{.}\end{aligned}$

But suppose we make both $\Delta H$ and $\Delta E$ one-tenth as large. Then the strip areas $\mathhlblue{E\cdot\Delta H}$ and $\mathhlgreen{H\cdot\Delta E}$ become as large, while the corner area $\mathhlviolet{\Delta H\cdot\Delta E}$ becomes as large. So this suggests as $\Delta H$ and $\Delta E$ get smaller and smaller, the corner shrinks much faster than either strip.

Now, following the approach we used in the previous lesson, imagine that these changes become infinitesimally small, replacing $\Delta H$ , $\Delta E$ , and $\Delta Q$ with the differentials $\dd{H}$ , $\dd{E}$ , and $\dd{Q}$ . Then we could imagine that the product of two infinitesimal changes $\mathhlviolet{\dd{H}\cdot\dd{E}}$ is compared with the terms $\mathhlblue{E\cdot\dd{H}}$ or $\mathhlgreen{H\cdot\dd{E}}$ , so we'll ignore the tiny corner.

This gives

$\boxed{\dd{Q}=E\dd{H}+H\dd{E}}\text{.}$

The first term, $E\dd{H}$ , is the contribution from changing the number of households, while the second, $H\dd{E}$ , is the contribution from changing the electricity use per household. Hence, this relationship now tells us how sensitive the total electricity consumption is to changes in either of these two factors.

It turns out that the regional planners have made projections for the growth of the metropolitan area over the next $$ 10 $$ years, letting $$ H $$ and $$ E $$ be functions of $$ t $$ , the number of years from the beginning of the projection.

First of all, their growth model suggests $H(t)=40+2t+0.05t^2\text{.}$ This model suggests that housing growth will . In fact, the rate at which the number of households changes is

$\displaystyle H'(t)=\)\(\displaystyle \text{,}$

which as time passes. (This rate is measured in .)

Meanwhile, because of improvements in efficiency and conservation, they project that $E(t)=12-0.5t\text{,}$ whose rate of change is

$\displaystyle E'(t)=\)\(\displaystyle \text{.}$

This suggests that average annual electricity use will as time passes. (This rate is measured in .)

Multiplying these together, we have $\begin{aligned}Q(t)&=H(t)\cdot E(t)\\ &=(40+2t+0.05t^2)(12-0.5t)\text{.}\end{aligned}$ Our model suggests that household growth pushes $$ Q $$ upward, while decreasing use per household pushes it downward. Therefore, from the signs alone, .

However, we can use what we now know about derivatives to figure out how $$ Q $$ changes with time.

Remember that we have two different ways of writing derivatives: $\dv{H}{t}=H'(t)\qand\dv{E}{t}=E'(t)\text{.}$

We can multiply through by $\dd{t}$ to give $\dd{H}=H'(t)\dd{t}\qand\dd{E}=E'(t)\dd{t}\text{.}$

Substituting these into our earlier expression for $\dd{Q}$ , we get

$\begin{aligned}\dd{Q}&= {E\cdot\dd{H}}+{H\cdot\dd{E}}\\ &={E(t)\cdot H'(t)\dd{t}}+{H(t)\cdot E'(t)\dd{t}}\\ &=\qty(\vphantom{\frac12}E(t)\cdot H'(t)+H(t)\cdot E'(t))\dd{t}\text{.}\end{aligned}$

Finally, dividing by $\dd{t}$ , we obtain

$\boxed{\dv{Q}{t}=E(t)\cdot H'(t)+H(t)\cdot E'(t)}\text{.}$

To see how this relationship works, consider the projection five years after the projection begins. We'll complete the values needed to determine how total use is changing:

Quantity	Value at $$ t=5 $$
$$ H(5) $$	thousand households
$$ E(5) $$	MWh per household
$$ H'(5) $$	thousand households per projection year
$$ E'(5) $$	MWh per household per projection year

The contribution from household growth is

$\displaystyle E(5)\cdot H'(5)=\)\(\displaystyle \text{ GWh per projection year,}$

while the contribution from changing per-household use is

$\displaystyle H(5)\cdot E'(5)=\)\(\displaystyle \text{ GWh per projection year.}$

Combining these contributions finally gives

$\displaystyle \eval{\dv{Q}{t}}_{t=5}=\)\(\displaystyle \text{ GWh per projection year.}$

At this point in the projection, the metropolitan area's total residential electricity use is therefore . This suggests that is the stronger of the two effects, although household growth substantially offsets the decrease.

Following a change through a function

In many real-world situations, several quantities depend on one another: a change in one quantity may change a second quantity, which then changes a third. To understand the overall effect, we need to follow the original change through each of these relationships.

Compositions of functions (that is, functions inside other functions) reflect these layers of dependence. When we recognize one expression nested inside another, we can give the inner expression a new variable name, separating the nested structure into simpler pieces. Then we can use differentials to tackle one layer at a time.

Introducing $$ u $$ didn't change the function — it simply gave the inner expression a shorter name so that we could focus on one layer at a time. Making this kind of substitution is often a useful problem-solving technique whenever one expression is nested inside another.

More generally, a composition $$ y=f(g(x)) $$ can be separated by letting $$ u=g(x) $$ and then writing $$ y=f(u) $$ . We first determine how $$ x $$ changes $$ u $$ , then use that differential to determine how $$ u $$ changes $$ y $$ . The general rule that describes this process is called the Chain Rule.

Theorem

Chain Rule

If $$ y=f(u) $$ and $$ u=g(x) $$ , then

$\dv{y}{x}=\dv{x}f(g(x))=f'(g(x))g'(x)\text.$

Equivalently,

$\dv{y}{x}=\dv{y}{u}\cdot\dv{u}{x}\text.$

Proof

Because $$ y=f(u) $$ and $$ u=g(x) $$ , their differentials are

$\dd{y}=f'(u)\dd{u}\qand\dd{u}=g'(x)\dd{x}\text{.}$

Substituting $$ u $$ and $\dd{u}$ , we get

$\dd{y}=f'(g(x))g'(x)\dd{x}\text.$

Dividing both sides by $\dd{x}$ gives

$\dv{y}{x}=f'(g(x))g'(x)\text.$

Alternatively, since $\displaystyle \dv{y}{u}=f'(u)$ and $\displaystyle \dv{u}{x}=g'(x)$ , we have

$\begin{aligned}\dv{y}{u}\cdot\dv{u}{x} &=f'(u)g'(x)\\ &=f'(g(x))g'(x)\text{.}\end{aligned}$

This is what we just got, so $\displaystyle \dv{y}{x}=\dv{y}{u}\cdot\dv{u}{x}$ .

Notice that the intermediate differential $\dd{u}$ appears to cancel; however, we haven't quite said what happens if $\dd{u}=0$ . Nevertheless the Chain Rule still works in this case, and we'll address this more carefully in a future section.

The Chain Rule suggests a practical way to differentiate nested functions: work through the function one layer at a time.

Start with the outermost function, differentiating that layer while leaving its input unchanged.
Move inward to differentiate the input.
Continue until the remaining differential is $\dd{x}$ (or whichever input is ultimately your independent variable).

Note

Working one layer at a time

The formula for the Chain Rule, $\displaystyle \dv{y}{x}=f'(g(x))g'(x)$ , could be applied to this example all at once, letting us write in a single step: $\dv{y}{x}=-2(x^2+3x+5)^{-3}(2x+3)\text{.}$

The differential calculation we did above organizes the same reasoning in a different way. We first write $\dd{y}=-2(x^2+3x+5)^{-3}\cdot\dd(x^2+3x+5)\text{,}$ temporarily leaving the inner differential unfinished. Then, after completing the outer layer, we return to $\dd(x^2+3x+5)$ and evaluate it. This lets us focus on one layer at a time instead of keeping every unfinished derivative in mind at once. The one-line form and the differential form produce the same result, but the layer-by-layer approach is often easier to manage, especially when a function has several nested layers.

Products of functions

The Explore activity showed that when the two factors in a product both change, the total change receives a contribution from each factor. We can use the same geometric argument to develop the Product Rule for any product $f(x)\cdot g(x)$ .

Theorem

Product Rule

If $$ f $$ and $$ g $$ are functions of $$ x $$ , then

$\begin{aligned}\dd(f\cdot g)&=g\cdot\dd{f}+f\cdot\dd{g}\text,\\ \dv{x}\bigl(f(x)\cdot g(x)\bigr)&=g(x)\cdot f'(x)+f(x)\cdot g'(x)\text.\end{aligned}$

Proof

We can visualize the product $f\cdot g$ as the area of a rectangle with side lengths $$ f $$ and $$ g $$ . If the sides increase by $\Delta f$ and $\Delta g$ , then the change in its area consists of two strips and one corner:

Loading the graph…If it does not appear, check the Desmos API configuration.

The original area is

f\cdot g

. Changing both side lengths adds two thin strips and a small corner.

Hence the change in area is $\Delta(fg)=g\Delta f+f\Delta g+\Delta f\Delta g\text.$

When the changes become infinitesimal, the corner $\dd{f}\dd{g}$ is a product of two infinitesimal changes, so we treat it as negligible compared with the two strips. This leaves

$\dd(f\cdot g)=g\cdot \dd{f}+f\cdot \dd{g}\text.$

If $$ f=f(x) $$ and $$ g=g(x) $$ , then $\dd{f}=f'(x)\dd{x}$ and $\dd{g}=g'(x)\dd{x}$ . Substituting these expressions and dividing by $\dd{x}$ gives the derivative form.

The differential form says to hold each factor fixed while the other changes, then add those two contributions. Notice that the derivative of a product is a sum of two contributions, not the product of the two derivatives.

The same idea extends to products with more factors: there is one contribution for each factor that changes. For example, if we have three factors, we get the following:

$\begin{aligned}\dd(fgh)&=(\dd{f})gh+f(\dd{g})h+fg(\dd{h})\text,\\ \dv{x}(f(x)g(x)h(x)) &= f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)\text.\end{aligned}$

Quotients of functions

We now know how to differentiate sums, differences, constant multiples, products, and compositions of functions. All that's left at this point is quotients. Luckily, a quotient can be rewritten as a product involving a negative power:

$\frac{f}{g}=f\cdot g^{-1}\text.$

This allows us to combine the Product Rule, the Power Rule, and the Chain Rule to develop a new formula called the Quotient Rule.

Theorem

Quotient Rule

If $g(x)\ne0$ , then

$\begin{aligned}\dd(\frac{f}{g})&=\frac{g\cdot\dd{f}-f\cdot\dd{g}}{g^2},\\ \dv{x}(\frac{f(x)}{g(x)})&=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{\big(g(x)\big)^2}\text{.}\end{aligned}$

Proof

First, the Product Rule gives $\dd(f\cdot g^{-1})=g^{-1}\cdot \dd{f}+f\cdot \dd(g^{-1})\text.$

For the second differential, the Power Rule gives us $\dd(g^{-1})=-g^{-2}\dd{g}$ , so we now have $\begin{aligned}\dd(f\cdot g^{-1}) &= g^{-1}\cdot \dd{f}+f\cdot \qty(-g^{-2}\dd{g})\\ &= g^{-1}\cdot \dd{f}-f\cdot g^{-2}\dd{g}\text.\end{aligned}$

Rewriting the negative powers as fractions again, we can finally combine them using a common denominator: $\begin{aligned}\dd(\frac{f}{g}) &=\frac{\dd{f}}{g}-\frac{f\dd{g}}{g^2}\\ &=\frac{g\cdot\dd{f}}{g^2}-\frac{f\cdot\dd{g}}{g^2}\\ &=\frac{g\cdot\dd{f}-f\cdot\dd{g}}{g^2}\text{.}\end{aligned}$

If $$ f $$ and $$ g $$ are functions of $$ x $$ , substitute $\dd{f}=f'(x)\dd{x}$ and $\dd{g}=g'(x)\dd{x}$ , then divide by $\dd{x}$ .

Note that in order for this argument to work, we do have to know that the Power Rule can be applied to negative exponents. We'll justify this in the next section.

The three rules in this lesson all follow the same working idea: write how one differential depends on others, substitute more differentials as necessary, and then divide by the differential of the independent variable when a derivative is desired.

Choosing a differentiation strategy

Before differentiating, look at how the expression is built:

Use the Chain Rule when one function is used as the input to another function.
Use the Product Rule when two or more nonconstant factors are multiplied.
Use the Quotient Rule when one nonconstant expression is divided by another.

More than one rule may be needed in the same problem. Work from the main operation connecting the largest pieces, then differentiate each piece using whatever rule its own structure requires.

Example

Find the derivative of $y=(3x-2)\sqrt{x^2+4}$ .

Solution

Since $$ y $$ is a product of two functions, we start with the Product Rule:

$\begin{aligned}\dd{y}&=\sqrt{x^2+4}\cdot\dd(3x-2)+(3x-2)\cdot\dd(\sqrt{x^2+4})\text{.}\end{aligned}$

Now we resolve the next two differentials: the first involves a linear function, and the second can be thought of as a power function with an exponent of $\tfrac12$ :

$\begin{aligned}\dd{y}&=\sqrt{x^2+4}\cdot(3\dd{x})+(3x-2)\cdot\frac{1}{2}(x^2+4)^{-1/2}\cdot\dd(x^2+4)\end{aligned}$

Finally we take care of the remaining differential and collect together the factors of $\dd{x}$ :

$\begin{aligned}\dd{y}&=3\sqrt{x^2+4}\cdot\dd{x}+(3x-2)\cdot\frac{1}{2}(x^2+4)^{-1/2}\cdot(2x\dd{x})\\ &=\qty(3\sqrt{x^2+4}+(3x-2)\frac{x}{\sqrt{x^2+4}})\dd{x}\text{.}\end{aligned}$

Dividing by $\dd{x}$ and cleaning the result up gives

$\dv{y}{x}=3\sqrt{x^2+4}+\frac{3x^2-2x}{\sqrt{x^2+4}}\text{.}$

However, the rule suggested by the original form is not always the easiest choice. By taking a moment to simplify the problem before differentiating, we can make things easier on ourselves.

Use the Chain Rule to find each derivative.
1. If $m(x)=(x^2+${c})^3$ , then $$ m'(x)= $$ .
  
  Solution
  
  Start by differentiating the outermost third power: $\dd{m}=3(x^2+${c})^2\cdot\dd(x^2+${c})$
  
  Then take care of the remaining differential: $\dd{m}=3(x^2+${c})^2\cdot2x\dd{x}$
  
  Finally, divide by $\dd{x}$ : $m'(x)=\dv{m}{x}=3(x^2+${c})^2\cdot2x$
2. If $f(x)=(${a}x^2-${b}x+${c})^4$ , then $$ f'(x)= $$ .
  
  Solution
  
  Start by differentiating $(\ \ )^4$ : $\dd{f}=4(${a}x^2-${b}x+${c})^3\cdot\dd(${a}x^2-${b}x+${c})$
  
  Then take care of the remaining differential: $\dd{f}=4(${a}x^2-${b}x+${c})^3\cdot(${2*a}x-${b})\dd{x}$
  
  Finally, divide by $\dd{x}$ : $f'(x)=\dv{f}{x}=4(${a}x^2-${b}x+${c})^3\cdot(${2*a}x-${b})$
3. If $g(x)=\sqrt{${a}x^3+${c}}$ , then $$ g'(x)= $$ .
  
  Solution
  
  Start by rewriting the square root as a power: $g(x)=(${a}x^3+${c})^{1/2}$
  
  Now differentiate $(\ \ )^{1/2}$ : $\dd{g}=\frac{1}{2}(${a}x^3+${c})^{-1/2}\cdot\dd(${a}x^3+${c})$
  
  Then take care of the remaining differential: $\dd{g}=\frac{1}{2}(${a}x^3+${c})^{-1/2}\cdot${3*a}x^2\dd{x}$
  
  Finally, divide by $\dd{x}$ and rewrite without negative powers: $g'(x)=\dv{g}{x}=\frac{${3*a}x^2}{2\sqrt{${a}x^3+${c}}}$
4. If $\displaystyle h(u)=\frac{1}{\sqrt{u^2+${c}}}$ , then $$ h'(u)= $$ .
  
  Solution
  
  Start by rewriting the reciprocal square root as a negative power: $h(u)=(u^2+${c})^{-1/2}$
  
  Now differentiate $(\ \ )^{-1/2}$ : $\dd{h}=-\frac{1}{2}(u^2+${c})^{-3/2}\cdot\dd(u^2+${c})$
  
  Then take care of the remaining differential: $\dd{h}=-\frac{1}{2}(u^2+${c})^{-3/2}\cdot2u\dd{u}$
  
  Finally, divide by $\dd{u}$ and simplify: $h'(u)=\dv{h}{u}=-\frac{u}{(u^2+${c})^{3/2}}$
5. If $k(t)=\qty(t^4+(${a}t-1)^3)^5$ , then $$ k'(t)= $$ .
  
  Solution
  
  Start with the outermost fifth power: $\dd{k}=5\qty(t^4+(${a}t-1)^3)^4\cdot\dd(t^4+(${a}t-1)^3)$
  
  Then differentiate the two terms in the remaining differential, leaving the innermost differential for later: $\dd{k}=5\qty(t^4+(${a}t-1)^3)^4\cdot\qty(4t^3\dd{t}+3(${a}t-1)^2\dd(${a}t-1))$
  
  Now take care of the final differential and collect $\dd{t}$ : $\dd{k}=5\qty(t^4+(${a}t-1)^3)^4\cdot\qty(4t^3+${3*a}(${a}t-1)^2)\dd{t}$
  
  Finally, divide by $\dd{t}$ : $k'(t)=\dv{k}{t}=5\qty(t^4+(${a}t-1)^3)^4\cdot\qty(4t^3+${3*a}(${a}t-1)^2)$
Use the Product Rule to find each derivative. (Do not multiply the factors out first.)
1. If $n(x)=(x+${a})(x^2+${b})$ , then $$ n'(x)= $$ .
  
  Solution
  
  Start by applying the Product Rule: $\dd{n}=(x^2+${b})\cdot\dd(x+${a})+(x+${a})\cdot\dd(x^2+${b})$
  
  Then take care of the two remaining differentials: $\dd{n}=(x^2+${b})\cdot\dd{x}+(x+${a})\cdot2x\dd{x}$
  
  Finally, collect $\dd{x}$ and divide by it: $n'(x)=\dv{n}{x}=(x^2+${b})+(x+${a})\cdot2x$
2. If $p(x)=(${a}x+${b})(x^2-${c})$ , then $$ p'(x)= $$ .
  
  Solution
  
  Start by applying the Product Rule: $\dd{p}=(x^2-${c})\cdot\dd(${a}x+${b})+(${a}x+${b})\cdot\dd(x^2-${c})$
  
  Then take care of the two remaining differentials: $\dd{p}=(x^2-${c})\cdot${a}\dd{x}+(${a}x+${b})\cdot2x\dd{x}$
  
  Finally, collect $\dd{x}$ and divide by it: $p'(x)=\dv{p}{x}=(x^2-${c})\cdot${a}+(${a}x+${b})\cdot2x$
3. If $q(x)=(x^2+${c})\sqrt{x}$ , then $$ q'(x)= $$ .
  
  Solution
  
  Start by applying the Product Rule: $\dd{q}=\sqrt{x}\cdot\dd(x^2+${c})+(x^2+${c})\cdot\dd(\sqrt{x})$
  
  Then take care of the two remaining differentials: $\dd{q}=\sqrt{x}\cdot2x\dd{x}+(x^2+${c})\cdot\frac{1}{2}x^{-1/2}\dd{x}$
  
  Finally, collect $\dd{x}$ and divide by it: $q'(x)=\dv{q}{x}=2x\cdot\sqrt{x}+\frac{x^2+${c}}{2\sqrt{x}}$
4. If $r(t)=t(1-t)(t^2+${c})$ , then $$ r'(t)= $$ .
  
  Solution
  
  Start with one contribution for each of the three factors: $\dd{r}=(\dd(t))\cdot(1-t)\cdot(t^2+${c})+t\cdot(\dd(1-t))\cdot(t^2+${c})+t\cdot(1-t)\cdot\dd(t^2+${c})$
  
  Then take care of the three remaining differentials: $\dd{r}=\dd{t}\cdot(1-t)\cdot(t^2+${c})+t\cdot(-1\dd{t})\cdot(t^2+${c})+t\cdot(1-t)\cdot2t\dd{t}$
  
  Now collect $\dd{t}$ : $\dd{r}=\qty((1-t)\cdot(t^2+${c})-t\cdot(t^2+${c})+t\cdot(1-t)\cdot2t)\dd{t}$
  
  Finally, divide by $\dd{t}$ : $r'(t)=\dv{r}{t}=(1-t)\cdot(t^2+${c})-t\cdot(t^2+${c})+t\cdot(1-t)\cdot2t$
5. If $s(u)=u^{-2}(${a}u^4-1)$ , then $$ s'(u)= $$ .
  
  Solution
  
  Start by applying the Product Rule: $\dd{s}=(${a}u^4-1)\cdot\dd(u^{-2})+u^{-2}\cdot\dd(${a}u^4-1)$
  
  Then take care of the two remaining differentials: $\dd{s}=(${a}u^4-1)\cdot(-2u^{-3})\dd{u}+u^{-2}\cdot${4*a}u^3\dd{u}$
  
  Finally, collect $\dd{u}$ , divide by it, and simplify: $s'(u)=\dv{s}{u}=(${a}u^4-1)\cdot(-2u^{-3})+${4*a}u$
Use the Quotient Rule to find each derivative.
1. If $\displaystyle w(x)=\frac{${a}x+${b}}{x+${c}}$ , then $$ w'(x)= $$ .
  
  Solution
  
  Start by applying the Quotient Rule: $\dd{w}=\frac{(x+${c})\cdot\dd(${a}x+${b})-(${a}x+${b})\cdot\dd(x+${c})}{(x+${c})^2}$
  
  Then take care of the two remaining differentials: $\dd{w}=\frac{(x+${c})\cdot${a}\dd{x}-(${a}x+${b})\cdot\dd{x}}{(x+${c})^2}$
  
  Now collect $\dd{x}$ : $\dd{w}=\frac{(x+${c})\cdot${a}-(${a}x+${b})}{(x+${c})^2}\dd{x}$
  
  Finally, divide by $\dd{x}$ : $w'(x)=\dv{w}{x}=\frac{(x+${c})\cdot${a}-(${a}x+${b})}{(x+${c})^2}$
2. If $\displaystyle a(x)=\frac{x^2+${b}}{x-${c}}$ , then $$ a'(x)= $$ .
  
  Solution
  
  Start by applying the Quotient Rule: $\dd{a}=\frac{(x-${c})\cdot\dd(x^2+${b})-(x^2+${b})\cdot\dd(x-${c})}{(x-${c})^2}$
  
  Then take care of the two remaining differentials: $\dd{a}=\frac{(x-${c})\cdot2x\dd{x}-(x^2+${b})\cdot\dd{x}}{(x-${c})^2}$
  
  Now collect $\dd{x}$ : $\dd{a}=\frac{(x-${c})\cdot2x-(x^2+${b})}{(x-${c})^2}\dd{x}$
  
  Finally, divide by $\dd{x}$ : $a'(x)=\dv{a}{x}=\frac{(x-${c})\cdot2x-(x^2+${b})}{(x-${c})^2}$
3. If $\displaystyle b(x)=\frac{\sqrt{x}}{x+${c}}$ , then $$ b'(x)= $$ .
  
  Solution
  
  Start by applying the Quotient Rule: $\dd{b}=\frac{(x+${c})\cdot\dd(\sqrt{x})-\sqrt{x}\cdot\dd(x+${c})}{(x+${c})^2}$
  
  Then take care of the two remaining differentials: $\dd{b}=\frac{(x+${c})\cdot\frac{1}{2}x^{-1/2}\dd{x}-\sqrt{x}\cdot\dd{x}}{(x+${c})^2}$
  
  Now collect $\dd{x}$ and rewrite without negative powers: $\dd{b}=\frac{\frac{x+${c}}{2\sqrt{x}}-\sqrt{x}}{(x+${c})^2}\dd{x}$
  
  Finally, divide by $\dd{x}$ : $b'(x)=\dv{b}{x}=\frac{\frac{x+${c}}{2\sqrt{x}}-\sqrt{x}}{(x+${c})^2}$
4. If $\displaystyle c(t)=\frac{${a}t-${b}}{t^2+${d}}$ , then $$ c'(t)= $$ .
  
  Solution
  
  Start by applying the Quotient Rule: $\dd{c}=\frac{(t^2+${d})\cdot\dd(${a}t-${b})-(${a}t-${b})\cdot\dd(t^2+${d})}{(t^2+${d})^2}$
  
  Then take care of the two remaining differentials: $\dd{c}=\frac{(t^2+${d})\cdot${a}\dd{t}-(${a}t-${b})\cdot2t\dd{t}}{(t^2+${d})^2}$
  
  Now collect $\dd{t}$ : $\dd{c}=\frac{(t^2+${d})\cdot${a}-(${a}t-${b})\cdot2t}{(t^2+${d})^2}\dd{t}$
  
  Finally, divide by $\dd{t}$ : $c'(t)=\dv{c}{t}=\frac{(t^2+${d})\cdot${a}-(${a}t-${b})\cdot2t}{(t^2+${d})^2}$
Mixed practice: Find each derivative. Decide which rule or combination of rules fits the structure of the function.
1. If $F(x)=(${a}x-${b})(x^2+${c})^4$ , then $$ F'(x)= $$ .
  
  Solution
  
  Start with the Product Rule: $\dd{F}=(x^2+${c})^4\cdot\dd(${a}x-${b})+(${a}x-${b})\cdot\dd((x^2+${c})^4)$
  
  Take care of the first differential and the outer layer of the second: $\dd{F}=(x^2+${c})^4\cdot${a}\dd{x}+(${a}x-${b})\cdot4(x^2+${c})^3\dd(x^2+${c})$
  
  Then take care of the remaining differential: $\dd{F}=(x^2+${c})^4\cdot${a}\dd{x}+(${a}x-${b})\cdot4(x^2+${c})^3\cdot2x\dd{x}$
  
  Now collect $\dd{x}$ : $\dd{F}=\qty((x^2+${c})^4\cdot${a}+(${a}x-${b})\cdot4(x^2+${c})^3\cdot2x)\dd{x}$
  
  Finally, divide by $\dd{x}$ : $F'(x)=\dv{F}{x}=(x^2+${c})^4\cdot${a}+(${a}x-${b})\cdot4(x^2+${c})^3\cdot2x$
2. If $\displaystyle G(x)=\qty(\frac{x^2+${c}}{${a}x-${b}})^3$ , then $$ G'(x)= $$ .
  
  Solution
  
  Start with the outermost third power: $\dd{G}=3\qty(\frac{x^2+${c}}{${a}x-${b}})^2\cdot\dd(\frac{x^2+${c}}{${a}x-${b}})$
  
  Then use the Quotient Rule on the remaining differential: $\dd{G}=3\qty(\frac{x^2+${c}}{${a}x-${b}})^2\cdot\frac{(${a}x-${b})\cdot\dd(x^2+${c})-(x^2+${c})\cdot\dd(${a}x-${b})}{(${a}x-${b})^2}$
  
  Now take care of the remaining differentials: $\dd{G}=3\qty(\frac{x^2+${c}}{${a}x-${b}})^2\cdot\frac{(${a}x-${b})\cdot2x\dd{x}-(x^2+${c})\cdot${a}\dd{x}}{(${a}x-${b})^2}$
  
  Then collect $\dd{x}$ : $\dd{G}=3\qty(\frac{x^2+${c}}{${a}x-${b}})^2\cdot\frac{(${a}x-${b})\cdot2x-(x^2+${c})\cdot${a}}{(${a}x-${b})^2}\dd{x}$
  
  Finally, divide by $\dd{x}$ : $G'(x)=\dv{G}{x}=3\qty(\frac{x^2+${c}}{${a}x-${b}})^2\cdot\frac{(${a}x-${b})\cdot2x-(x^2+${c})\cdot${a}}{(${a}x-${b})^2}$
3. If $\displaystyle H(x)=\frac{x^2+${c}}{\sqrt{${a}x+${b}}}$ , then $$ H'(x)= $$ .
  
  Solution
  
  Start by rewriting the denominator as a negative power: $H(x)=(x^2+${c})\cdot(${a}x+${b})^{-1/2}$
  
  Then apply the Product Rule: $\dd{H}=(${a}x+${b})^{-1/2}\cdot\dd(x^2+${c})+(x^2+${c})\cdot\dd((${a}x+${b})^{-1/2})$
  
  Take care of the first differential and the outer layer of the second: $\dd{H}=(${a}x+${b})^{-1/2}\cdot2x\dd{x}+(x^2+${c})\cdot\qty(-\frac{1}{2})(${a}x+${b})^{-3/2}\dd(${a}x+${b})$
  
  Then take care of the remaining differential: $\dd{H}=(${a}x+${b})^{-1/2}\cdot2x\dd{x}+(x^2+${c})\cdot\qty(-\frac{1}{2})(${a}x+${b})^{-3/2}\cdot${a}\dd{x}$
  
  Now collect $\dd{x}$ : $\dd{H}=\qty((${a}x+${b})^{-1/2}\cdot2x+(x^2+${c})\cdot\qty(-\frac{1}{2})(${a}x+${b})^{-3/2}\cdot${a})\dd{x}$
  
  Finally, divide by $\dd{x}$ : $H'(x)=\dv{H}{x}=(${a}x+${b})^{-1/2}\cdot2x+(x^2+${c})\cdot\qty(-\frac{1}{2})(${a}x+${b})^{-3/2}\cdot${a}$
4. If $\displaystyle J(x)=\frac{(x^2+${a})(x^3-${b})}{${c}x+${d}}$ , then $$ J'(x)= $$ .
  
  Solution
  
  Start with the Quotient Rule, leaving the numerator differential for later: $\dd{J}=\frac{(${c}x+${d})\cdot\dd((x^2+${a})(x^3-${b}))-(x^2+${a})\cdot(x^3-${b})\cdot\dd(${c}x+${d})}{(${c}x+${d})^2}$
  
  Then apply the Product Rule inside the numerator: $\dd((x^2+${a})(x^3-${b}))=(x^3-${b})\cdot\dd(x^2+${a})+(x^2+${a})\cdot\dd(x^3-${b})$
  
  Take care of the differentials in that Product Rule: $\dd((x^2+${a})(x^3-${b}))=(x^3-${b})\cdot2x\dd{x}+(x^2+${a})\cdot3x^2\dd{x}$
  
  Also, $\dd(${c}x+${d})=${c}\dd{x}$
  
  Substitute these results into the Quotient Rule: $\dd{J}=\frac{(${c}x+${d})\cdot\qty((x^3-${b})\cdot2x\dd{x}+(x^2+${a})\cdot3x^2\dd{x})-(x^2+${a})\cdot(x^3-${b})\cdot${c}\dd{x}}{(${c}x+${d})^2}$
  
  Now collect $\dd{x}$ : $\dd{J}=\frac{(${c}x+${d})\cdot\qty((x^3-${b})\cdot2x+(x^2+${a})\cdot3x^2)-(x^2+${a})\cdot(x^3-${b})\cdot${c}}{(${c}x+${d})^2}\dd{x}$
  
  Finally, divide by $\dd{x}$ : $J'(x)=\dv{J}{x}=\frac{(${c}x+${d})\cdot\qty((x^3-${b})\cdot2x+(x^2+${a})\cdot3x^2)-(x^2+${a})\cdot(x^3-${b})\cdot${c}}{(${c}x+${d})^2}$

The table gives selected values of two functions $$ f $$ and $$ g $$ .

$$ x $$	$$ f(x) $$	$$ f'(x) $$	$$ g(x) $$	$$ g'(x) $$
$$ 1 $$	$$ ${f1} $$	$$ ${fp1} $$	$$ ${g1} $$	$$ ${gp1} $$
$$ 2 $$	$$ ${f2} $$	$$ ${fp2} $$	$$ ${g2} $$	$$ ${gp2} $$
$$ 3 $$	$$ ${f3} $$	$$ ${fp3} $$	$$ ${g3} $$	$$ ${gp3} $$

Use the table to find each value:

Let $$ P(x)=f(x)g(x) $$ . Find $P(${p_x})$ and $P'(${p_x})$ .

$P(${p_x})=$ and $P'(${p_x})=$ .

Solution

First of all, we have $\begin{aligned}P(${p_x})&=f(${p_x})\cdot g(${p_x})\\ &=${p_f}\cdot ${p_g}\\ &=${p_value}\text.\end{aligned}$

Then, using the Product Rule, we have $\begin{aligned}P'(${p_x})&=g(${p_x})\cdot f'(${p_x})+f(${p_x})\cdot g'(${p_x})\\ &=${p_g}\cdot(${p_fp})+${p_f}\cdot(${p_gp})\\ &=${p_derivative}\text.\end{aligned}$
Let $\displaystyle R(x)=\frac{f(x)}{g(x)}$ . Find $R(${r_x})$ and $R'(${r_x})$ .

$R(${r_x})=$ and $R'(${r_x})=$ .

Solution

First of all, we have $\begin{aligned}R(${r_x})&=\frac{f(${r_x})}{g(${r_x})}\\ &=\frac{${r_f}}{${r_g}}\\ &=${r_value}\text.\end{aligned}$

Then, using the Quotient Rule, we have $\begin{aligned}R'(${r_x})&=\frac{g(${r_x})\cdot f'(${r_x})-f(${r_x})\cdot g'(${r_x})}{g(${r_x})^2}\\ &=\frac{${r_g}\cdot(${r_fp})-${r_f}\cdot(${r_gp})}{${r_g}^2}\\ &=${r_derivative}\text.\end{aligned}$
Let $$ C(x)=f(g(x)) $$ . Find $C(${c_x})$ and $C'(${c_x})$ .

$C(${c_x})=$ and $C'(${c_x})=$ .

Solution

First of all, the table tells us that $g(${c_x})=${c_g}$ . Therefore, $\begin{aligned}C(${c_x})&=f(g(${c_x}))\\ &=f(${c_g})\\ &=${c_value}\text.\end{aligned}$

Then, using the Chain Rule, we have $\begin{aligned}C'(${c_x})&=f'(g(${c_x}))\cdot g'(${c_x})\\ &=f'(${c_g})\cdot(${c_gp})\\ &=(${c_fp_at_g})\cdot(${c_gp})\\ &=${c_derivative}\text.\end{aligned}$

The graph shows two linear functions, $$ f $$ and $$ g $$ , on the interval $0\leq x\leq${d}$ . Use values and slopes read from the graph to find each derivative.

Loading the graph…If it does not appear, check the Desmos API configuration.
1. Let $$ P(x)=f(x)g(x) $$ . Find $P'(${p_x})$ .
  
  $P'(${p_x})=$ .
  
  Solution
  
  From the graph, $f(${p_x})=${p_f}$ and $g(${p_x})=${p_g}$ . Since the functions are linear, their slopes are constant: $f'(${p_x})=${m}$ and $g'(${p_x})=-1$ .
  
  Using the Product Rule, $\begin{aligned}P'(${p_x})&=g(${p_x})\cdot f'(${p_x})+f(${p_x})\cdot g'(${p_x})\\ &=${p_g}\cdot${m}+(${p_f})\cdot(-1)\\ &=${p_answer}\text.\end{aligned}$
2. Let $\displaystyle R(x)=\frac{f(x)}{g(x)}$ . Find $R'(${r_x})$ .
  
  $R'(${r_x})=$ .
  
  Solution
  
  From the graph, $f(${r_x})=${r_f}$ and $g(${r_x})=${r_g}$ . The slopes give $f'(${r_x})=${m}$ and $g'(${r_x})=-1$ .
  
  Using the Quotient Rule, $\begin{aligned}R'(${r_x})&=\frac{g(${r_x})\cdot f'(${r_x})-f(${r_x})\cdot g'(${r_x})}{g(${r_x})^2}\\ &=\frac{${r_g}\cdot${m}-(${r_f})\cdot(-1)}{${r_g}^2}\\ &=${r_answer}\text.\end{aligned}$
3. Let $$ C(x)=f(g(x)) $$ . Find $C'(${c_x})$ .
  
  $C'(${c_x})=$ .
  
  Solution
  
  First read $g(${c_x})=${c_g}$ from the graph. At the resulting input, the slope of $$ f $$ is $f'(${c_g})=${m}$ . The slope of $$ g $$ is $g'(${c_x})=-1$ .
  
  Using the Chain Rule, $\begin{aligned}C'(${c_x})&=f'(g(${c_x}))\cdot g'(${c_x})\\ &=f'(${c_g})\cdot(-1)\\ &=${m}\cdot(-1)\\ &=${c_answer}\text.\end{aligned}$
Suppose a car is traveling at a speed of $$ v $$ miles per hour when the driver notices a hazard. Its stopping distance $$ L $$ , measured in car lengths, can be modeled by

$\[L(v)=0.0034v^2+0.15v.\]$

This model uses an average car length of $$ 14.7 $$ feet. Now suppose that the car's speed $$ v(t) $$ depends on time $$ t $$ , measured in seconds.
1. First find the differential $\dd{L}$ in terms of $$ v $$ and $\dd{v}$ .
  
  $\dd{L}=$ $\dd{v}$ .
  
  Solution
  
  Differentiate the stopping-distance model with respect to speed:
  
  $\begin{aligned}\dd{L}&=\dd(0.0034v^2+0.15v)\\ &=0.0068v\dd{v}+0.15\dd{v}\\ &=\qty(0.0068v+0.15)\dd{v}.\end{aligned}$
2. At an instant when the car is traveling at $$ 50 $$ miles per hour, how sensitive is stopping distance to speed?
  
  $\displaystyle \eval{\dv{L}{v}}_{v=50}=$ car lengths for each additional mile per hour.
  
  Solution
  
  From the previous part,
  
  $\begin{aligned}\eval{\dv{L}{v}}_{v=50}&=0.0068(50)+0.15\\ &=0.49\text{ car lengths for each additional mile per hour.}\end{aligned}$
  
  Near $$ 50 $$ miles per hour, each additional mile per hour adds about $$ 0.49 $$ car lengths to the predicted stopping distance.
3. At that same instant, suppose the car's speed is increasing at $$ 4 $$ miles per hour per second. At what rate is its predicted stopping distance changing?
  
  $\displaystyle \dv{L}{t}=$ car lengths per second.
  
  Solution
  
  The change in time first produces a change in speed:
  
  $\dd{v}=4\dd{t}.$
  
  At $$ v=50 $$ , we already found that $\dd{L}=0.49\dd{v}$ . Substituting the change in speed gives
  
  $\begin{aligned}\dd{L}&=0.49\dd{v}\\ &=0.49\cdot4\dd{t}\\ &=1.96\dd{t}.\end{aligned}$
  
  Therefore, $\dv{L}{t}=1.96$ car lengths per second. At this instant, the predicted stopping distance is increasing by about $$ 1.96 $$ car lengths each second.
Model adapted from The Physics Behind Stopping a Car by P. Lutus.
A solar farm's total rated generating capacity is

$\[G(t)=N(t)P(t),\]$

where $$ N(t) $$ is the number of installed panels and $$ P(t) $$ is the average rated capacity of a panel, in kilowatts. At a certain time, the solar farm has $$ 12{,}000 $$ panels with an average rated capacity of $$ 0.4 $$ kilowatts per panel. The number of panels is increasing by $$ 800 $$ panels per year, while the average rated capacity is increasing by $$ 0.01 $$ kilowatts per panel per year.
1. Find the contribution to $$ G'(t) $$ from installing additional panels.
  
  kilowatts per year.
  
  Solution
  
  The contribution from changing the number of panels is $P(t)\cdot N'(t)$ . Therefore,
  
  $P(t)\cdot N'(t)=0.4\cdot800=320\text{ kilowatts per year.}$
2. Find the contribution to $$ G'(t) $$ from the increasing average rated capacity of a panel.
  
  kilowatts per year.
  
  Solution
  
  The contribution from changing the average rated capacity is $N(t)\cdot P'(t)$ . Therefore,
  
  $N(t)\cdot P'(t)=12{,}000\cdot0.01=120\text{ kilowatts per year.}$
3. At what rate is the solar farm's total rated generating capacity changing?
  
  $$ G'(t)= $$ kilowatts per year.
  
  Solution
  
  Using the Product Rule,
  
  $\begin{aligned}G'(t)&=P(t)\cdot N'(t)+N(t)\cdot P'(t)\\ &=0.4\cdot800+12{,}000\cdot0.01\\ &=320+120\\ &=440\text{ kilowatts per year.}\end{aligned}$
  
  Both changes increase the total rated capacity, so their contributions add.
A manufacturer has a total cost of $$ C(x) $$ dollars when it produces $$ x $$ units. Its average cost per unit is

$A(x)=\frac{C(x)}{x}.$
1. Use the Quotient Rule to find $$ A'(x) $$ in terms of $$ C(x) $$ and $$ C'(x) $$ .
  
  Choose one $\displaystyle A'(x)=\frac{x\cdot C'(x)-C(x)}{x^2}$ $\displaystyle A'(x)=\frac{C'(x)}{x}$ $\displaystyle A'(x)=\frac{C(x)-x\cdot C'(x)}{x^2}$ $\displaystyle A'(x)=\frac{x\cdot C'(x)-C(x)}{x}$
  
  Solution
  
  Apply the Quotient Rule with numerator $$ C(x) $$ and denominator $$ x $$ :
  
  $\begin{aligned}\dd{A}&=\frac{x\cdot\dd(C(x))-C(x)\cdot\dd{x}}{x^2}\\ &=\frac{x\cdot C'(x)\dd{x}-C(x)\cdot\dd{x}}{x^2}\\ &=\frac{x\cdot C'(x)-C(x)}{x^2}\dd{x}.\end{aligned}$
  
  Dividing by $\dd{x}$ gives $A'(x)=\frac{x\cdot C'(x)-C(x)}{x^2}.$
2. Suppose $$ C(200)=14{,}000 $$ and $$ C'(200)=50 $$ . Find the average cost and its rate of change when $$ 200 $$ units are produced.
  
  $A(200)=\$$ per unit and $$ A'(200)= $$ dollars per unit for each additional unit produced.
  
  Solution
  
  First, the average cost is
  
  $A(200)=\frac{C(200)}{200}=\frac{14{,}000}{200}=70\text{ dollars per unit.}$
  
  Then,
  
  $\begin{aligned}A'(200)&=\frac{200\cdot C'(200)-C(200)}{200^2}\\ &=\frac{200\cdot50-14{,}000}{200^2}\\ &=-0.1\text{ dollars per unit for each additional unit produced.}\end{aligned}$
3. Which statement best interprets $$ A'(200)=-0.1 $$ ?
  
  Choose one Near a production level of $$ 200 $$ units, producing one additional unit decreases the average cost by about $\$0.10$ per unit.Producing $$ 200 $$ units costs a total of $\$0.10$ .Each of the first $$ 200 $$ units costs $\$0.10$ to produce.Near a production level of $$ 200 $$ units, producing one additional unit decreases total cost by about $\$0.10$ .
  
  Solution
  The derivative $$ A'(200) $$ describes how the average cost per unit changes as the number of units produced changes. Its negative sign means that average cost is decreasing near $$ x=200 $$ .
Let $\displaystyle L(x)=\frac{${a}x+${b}}{x^{${c}}}$ . Find $$ L'(x) $$ in each of three ways. Leave each intermediate answer in the form naturally produced by the indicated method.
1. Use the Quotient Rule:
  
  $$ L'(x)= $$ .
  
  Solution
  
  Start with the Quotient Rule: $\dd{L}=\frac{x^{${c}}\cdot\dd(${a}x+${b})-(${a}x+${b})\cdot\dd(x^{${c}})}{x^{${2*c}}}$
  
  Then take care of the two remaining differentials: $\dd{L}=\frac{x^{${c}}\cdot${a}\dd{x}-(${a}x+${b})\cdot${c}x^{${c-1}}\dd{x}}{x^{${2*c}}}$
  
  Now collect $\dd{x}$ : $\dd{L}=\frac{x^{${c}}\cdot${a}-(${a}x+${b})\cdot${c}x^{${c-1}}}{x^{${2*c}}}\dd{x}$
  
  Finally, divide by $\dd{x}$ : $L'(x)=\frac{x^{${c}}\cdot${a}-(${a}x+${b})\cdot${c}x^{${c-1}}}{x^{${2*c}}}$
2. Rewrite $L(x)=(${a}x+${b})\cdot x^{-${c}}$ and use the Product Rule:
  
  $$ L'(x)= $$ .
  
  Solution
  
  Start with the Product Rule: $\dd{L}=x^{-${c}}\cdot\dd(${a}x+${b})+(${a}x+${b})\cdot\dd(x^{-${c}})$
  
  Then take care of the two remaining differentials: $\dd{L}=x^{-${c}}\cdot${a}\dd{x}+(${a}x+${b})\cdot(-${c})x^{-${c+1}}\dd{x}$
  
  Now collect $\dd{x}$ : $\dd{L}=\qty(x^{-${c}}\cdot${a}+(${a}x+${b})\cdot(-${c})x^{-${c+1}})\dd{x}$
  
  Finally, divide by $\dd{x}$ : $L'(x)=x^{-${c}}\cdot${a}+(${a}x+${b})\cdot(-${c})x^{-${c+1}}$
3. Distribute and simplify $$ L(x) $$ before differentiating:
  
  $$ L'(x)= $$ .
  
  Solution
  
  Start by distributing the denominator and simplifying: $L(x)=${a}x^{${1-c}}+${b}x^{-${c}}$
  
  Then differentiate the two terms: $\dd{L}=${a}\cdot(${1-c})x^{-${c}}\dd{x}+${b}\cdot(-${c})x^{-${c+1}}\dd{x}$
  
  Now collect $\dd{x}$ and simplify the coefficients: $\dd{L}=\qty(${a*(1-c)}x^{-${c}}-${b*c}x^{-${c+1}})\dd{x}$
  
  Finally, divide by $\dd{x}$ : $L'(x)=${a*(1-c)}x^{-${c}}-${b*c}x^{-${c+1}}$
Let $$ c $$ be a constant and let $F(x)=c\cdot f(x)$ .
1. Because $$ c $$ does not change, what is $\dd{c}$ ?
  
  $\dd{c}=$ .
  
  Solution
  A constant has no change, so $\dd{c}=0$ .
2. Apply the Product Rule to $F=c\cdot f$ . Which term disappears when $\dd{c}=0$ ?
  
  Choose one $f\cdot\dd{c}$ $c\cdot\dd{f}$ Both termsNeither term
  
  Solution
  
  The Product Rule gives $\dd{F}=f\cdot\dd{c}+c\cdot\dd{f}.$
  
  Since $\dd{c}=0$ , the first term disappears:
  
  $\dd{F}=f\cdot0+c\cdot\dd{f}=c\cdot\dd{f}.$
  
  Dividing by $\dd{x}$ gives $F'(x)=c\cdot f'(x),$ which is exactly the Constant Multiple Rule. Thus, the Constant Multiple Rule is the special case of the Product Rule in which one factor is constant.
Let $P(x)=f(x)\cdot g(x)\cdot h(x)$ .
1. First view $$ P $$ as the product $\qty(f\cdot g)\cdot h$ . Which equation follows from the Product Rule?
  
  Choose one $\dd{P}=h\cdot\dd(f\cdot g)+f\cdot g\cdot\dd{h}$ $\dd{P}=f\cdot g\cdot\dd{h}$ $\dd{P}=g\cdot h\cdot\dd{f}+f\cdot h\cdot\dd{g}$ $\dd{P}=f\cdot\dd{g}+g\cdot\dd{h}+h\cdot\dd{f}$
  
  Solution
  
  Regarding $f\cdot g$ as one factor and $$ h $$ as the other, the Product Rule gives
  
  $\dd{P}=h\cdot\dd(f\cdot g)+f\cdot g\cdot\dd{h}.$
2. Now apply the Product Rule to the remaining differential $\dd(f\cdot g)$ . Which formula results after distributing $$ h $$ ?
  
  Choose one $\dd{P}=\dd{f}\cdot g\cdot h+f\cdot\dd{g}\cdot h+f\cdot g\cdot\dd{h}$ $\dd{P}=f\cdot\dd{g}+g\cdot\dd{f}+h\cdot\dd{h}$ $\dd{P}=h\cdot\dd{f}+h\cdot\dd{g}+f\cdot g\cdot\dd{h}$ $\dd{P}=f\cdot g\cdot h\cdot\qty(\dd{f}+\dd{g}+\dd{h})$
  
  Solution
  
  First replace $\dd(f\cdot g)$ with $g\cdot\dd{f}+f\cdot\dd{g}$ :
  
  $\begin{aligned}\dd{P}&=h\cdot\dd(f\cdot g)+f\cdot g\cdot\dd{h}\\ &=h\cdot\qty(g\cdot\dd{f}+f\cdot\dd{g})+f\cdot g\cdot\dd{h}\\ &=\dd{f}\cdot g\cdot h+f\cdot\dd{g}\cdot h+f\cdot g\cdot\dd{h}.\end{aligned}$
  
  If all three functions depend on $$ x $$ , then substituting $\dd{f}=f'(x)\dd{x}$ , $\dd{g}=g'(x)\dd{x}$ , and $\dd{h}=h'(x)\dd{x}$ and dividing by $\dd{x}$ gives
  
  $P'(x)=f'(x)\cdot g(x)\cdot h(x)+f(x)\cdot g'(x)\cdot h(x)+f(x)\cdot g(x)\cdot h'(x).$
  
  Each term differentiates one factor while leaving the other two unchanged.
In this exercise, you'll use the Product Rule to explain why the Power Rule works for every positive integer exponent.
1. The pattern begins with $$ x^1=x $$ . What is $\dd(x^1)$ ?
  
  $\dd(x^1)=$ $\dd{x}$ .
  
  Solution
  
  Since $$ x^1=x $$ ,
  
  $\dd(x^1)=\dd{x}=1\cdot x^0\dd{x},$
  
  which agrees with the Power Rule when the exponent is $$ 1 $$ .
2. Suppose the Power Rule already works for some positive integer $$ n $$ , so that $\dd(x^n)=n\cdot x^{n-1}\dd{x}$ . Rewrite $x^{n+1}=x^n\cdot x$ and apply the Product Rule. Which calculation shows that the rule must then work for the next exponent, $$ n+1 $$ ?
  
  Choose one $\dd(x^{n+1})=x\cdot n\cdot x^{n-1}\dd{x}+x^n\dd{x}=(n+1)\cdot x^n\dd{x}$ $\dd(x^{n+1})=n\cdot x^{n-1}\dd{x}+\dd{x}=n\cdot x^n\dd{x}$ $\dd(x^{n+1})=x^n\cdot n\cdot x^{n-1}\dd{x}=n\cdot x^{2n-1}\dd{x}$ $\dd(x^{n+1})=(n+1)\cdot x^{n+1}\dd{x}$
  
  Solution
  
  Using $x^{n+1}=x^n\cdot x$ , the Product Rule gives
  
  $\begin{aligned}\dd(x^{n+1})&=x\dd(x^n)+x^n\dd{x}\\ &=x\cdot n\cdot x^{n-1}\dd{x}+x^n\dd{x}\\ &=n\cdot x^n\dd{x}+x^n\dd{x}\\ &=(n+1)\cdot x^n\dd{x}.\end{aligned}$
  
  This is precisely the Power Rule for the exponent $$ n+1 $$ .
3. Why does this establish the Power Rule for every positive integer exponent?
  
  Choose one The rule works for $$ x^1 $$ , and whenever it works for one positive integer exponent, the Product Rule shows that it works for the next one.The rule has now been checked separately for every positive integer.Every positive integer exponent is equal to $$ 1 $$ .The Product Rule applies only when the exponent is positive.
  
  Solution
  We know the pattern works for $$ x^1 $$ . The calculation in the previous part carries the pattern from $$ x^1 $$ to $$ x^2 $$ , then from $$ x^2 $$ to $$ x^3 $$ , and so on. Repeating the same argument reaches every positive integer exponent.

$\( x \)$	$\( f(x) \)$	$\( f'(x) \)$	$\( g(x) \)$	$\( g'(x) \)$
$\( 1 \)$	$\( ${f1} \)$	$\( ${fp1} \)$	$\( ${g1} \)$	$\( ${gp1} \)$
$\( 2 \)$	$\( ${f2} \)$	$\( ${fp2} \)$	$\( ${g2} \)$	$\( ${gp2} \)$
$\( 3 \)$	$\( ${f3} \)$	$\( ${fp3} \)$	$\( ${g3} \)$	$\( ${gp3} \)$

Quantity	Value at $\( t=5 \)$
$\( H(5) \)$	thousand households
$\( E(5) \)$	MWh per household
$\( H'(5) \)$	thousand households per projection year
$\( E'(5) \)$	MWh per household per projection year