Chapter 12The Basic Arithmetic Of Calculus

Remember learning arithmetic?After seeing how to multiply small numbers, we learned how to multiply numbers with several digits:

\[ 13 \times 15 = (10 + 3)(10 + 5) = 100 + 30 + 50 + 15 \]

We can’t just combine the first digits (10$ \times $10) and the second (3$ \times $5) and call it done.We have to walk through the cross-multiplication.

微积分是相似的。If we have the whole function, we can blithely say that$ f(x) $has derivative$ f'(x) $.But that isn’t illuminating, or explaining what happens behind the scenes.

If we can describe our function in terms of a building block$ x $(such as$ f(x) = 3x^2 + x $), then we should be able to find the derivative, the pattern of changes, in terms of that same building block.If we have two types of building blocks ($ f = a \cdot b $，我们就会得到这两个构造块的导数。

Here’s the general strategy:

Imagine a scenario with a few building blocks ($ area = length \cdot width $)
Let every component change
Measure the change in the overall system
Remove the measurement artifacts (our instruments interfering with each other)

一旦我们知道系统是如何分解的，我们就可以将它们逆向工程到积分中(FTOC太棒了!)

12.1Addition

Let’s start off easy: how does a system with two added components behave?

In the real world, this could be sending two friends (Frank and George) to build a fence.Let’s say Frank gets the wood, and George gets the paint.总费用是多少?

\[ \textit{Total} = \textit{Frank's cost} + \textit{George's cost} \]

\[ t(x) = f(x) + g(x) \]

The derivative of the entire system,$ \frac{dt}{dx} $, is the cost per additional foot.Intuitively, we suspect the total increase is the sum of the increases in the parts:

\[ \frac{dt}{dx} = \frac{df}{dx} + \frac{dg}{dx} \]

That relationship makes sense, right?Let’s say Frank’s cost is \$3/foot for the wood, and George adds \$0.50/foot for the paint.If we ask for another foot, the total cost will increase by \$3.50.

Here’s the math for that result:

Original:$ f + g $
New:$ (f + df) + (g + dg) $
Change:$ (f + df) + (g + dg) - (f + g) = df + dg $

In my head, I imagine$ x $，你要的钱，在一个角落里默默地变着。This creates a visible change in$ f $(size$ df $) and$ g $(size$ dg $), and we see the total change as$ df + dg $.

It seems we should just combine the total up front, writing$ total = 3.5x $not$ total = f(x) + g(x) = 3x + 0.5x $.Normally, we would simplify an equation, but it’s sometimes helpful to list every contribution (total = base + shipping + tax).就我们的情况而言，我们看到弗兰克对价格贡献最大。

Remembering the derivative is the “per$ dx $” rate, we write:

\[ \frac{d}{dx} \left( f(x) + g(x) \right) = \frac{df}{dx} + \frac{dg}{dx} \]

但是，看看这些符号!Let’s trim it down:

Write$ f $instead of$ f(x) $.We’ll assume a single letter is an entire function, and by the Third Edict of The Grand Math Poombahs, our functions will use a parameter$ x $.
We’ll express the derivative using a single quote ($ f' $), not with a ratio ($ \frac{df}{dx} $).We’re most interested in the relationship between the parts (addition), not the gritty details of the parts themselves.

So now the addition rule becomes:

\[ (f + g)' = f' + g' \]

Much better!Here’s how I read it: Take a system made of several parts:$ (f + g) $.The change in the overall system,$ (f + g)' $, can be found by adding the change from each part.

12.2Multiplication

Let’s try a tricker scenario.Instead of inputs that are added (almost oblivious to each other), what if they are multiplied?

Suppose Frank and George are making arectangular为你的花园。Frank handles the width and George takes care of the height.Whenever you clap, they move… but by different amounts!

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Frank’s steps are 3-feet long, but George’s are only 2-feet long (zookeeping accident, don’t ask).How can we describe the system?

\[ \textit{Area} = \textit{width} \cdot \textit{height} = f(x) \cdot g(x) \]

\[ f(x) = 3x \]

\[ g(x) = 2x \]

We have linear parts, so the derivatives are simple:$ f'(x) = 3 $and$ g'(x) = 2 $.下一次鼓掌会发生什么?

//www.i494.com/wp-content/uploads/calculus/course/lesson7/garden-change

Looks familiar!We have a horizontal strip, a vertical strip, and a corner piece.We can work out the amounts with algebra:

Original:$ f \cdot g $
New:$ (f + df) \cdot (g + dg) = (f\cdot g) + (f\cdot dg) + (g\cdot df) + (df\cdot dg) $
Change:$ f\cdot dg + g\cdot df + df\cdot dg $

Let’s see this change more closely:

The horizontal strip happened when$ f $changed (by$ df $), and$ g $was the same value
The vertical strip was made when$ g $changed (by$ dg $), and$ f $was the same value
The corner piece ($ df \cdot dg $) happened when the change in one component ($ df $) interacted with the change in the other ($ dg $)

The corner piece is our sample measurement getting tangled on itself, and should be removed.(If we’reforcedto move in whole units, then the corner is fine.But most real-world systems can change continuously, by any decimal number, and we want the measurement artifacts removed.)

To find the total change, we drop the$ df \cdot dg $term (interference between the changes) and get:

\[ f \cdot dg + g \cdot df \]

I won’t let you forget the derivative is on a “per dx” basis, so we write:

\[ \frac{\textit{total change}}{dx} = f \frac{dg}{dx} + g \frac{df}{dx} \]

\[ (f \cdot g)' = f \cdot g' + g \cdot f' \]

There is an implicit “x” changing off in the distance, which makes$ f $and$ g $move.We hide these details to make the notation simpler.

用英语来说就是:假设有多个部分的情况。As they change, and continue to be multiplied, add up the new horizontal and vertical strips that are formed.

Let’s try out the rule: if we have a 12$ \times $8 garden and increment by a whole step, what change will we see?

In this case, we’ll use the discrete version of the rule since we’re forced to move as a whole step:

Vertical strip:$ f \cdot dg = 12 \cdot 2 = 24 $
Horizontal strip$ g \cdot df = 8 \cdot 3 = 24 $
Corner piece:$ df \cdot dg = 3 \cdot 2 = 6 $
Total change:$ 24 + 24 + 6 = 54 $

让我们来测试一下。We go from 12$ \times $8(96平方英尺)至15$ \times $10 (150 square feet).And yep, the area increase was 150 - 96 = 54 square feet!

12.3Simple Division (Inverses)

Inverses can be tough to visualize: as$ x $gets bigger,$ \frac{1}{x} $gets smaller.Let’s take it slow.

Suppose you’re sharing a cake with Frank.You’ve just cut it in half, about to take a bite and… George shuffles in.He looks upset, and you’re not about to mention the fresh set of claw marks.

But you’ve just cut the cake in half, what can you do?

切一遍。You and Frank can cut your existing portion in thirds, and give George a piece:

//www.i494.com/wp-content/uploads/calculus/course/lesson7/cake-cut

Neat!现在每个人都有总数的1/3。You gave up 1/3 of your amount (1/2), that is, you each gave George 1/6 of the total.

Time to eat!But just as you’re about to bite in… the veggie godfather walks in.Oh, he’lldefinitelywant a piece.What do you do?

切一遍。Everyone smooshes together their portion, cuts it infourths, and hands one piece to the Don.The cake is split evenly again.

This is step-by-step thinking applied to division:

Your original share is$ \frac{1}{x} $(when x=2, you have 1/2)
Someone walks in
Your new share becomes$ \frac{1}{x + 1} $

你的蛋糕量是怎么变化的?Well, you took your original slice ($ \frac{1}{x} $), cut it into the new number of pieces ($ \frac{1}{x + 1} $), and gave one away (the change is negative):

\[ \frac{1}{x} \cdot \frac{-1}{x + 1} = \frac{-1}{x(x + 1)} \]

We can probably guess that the +1 is a measurement artifact because we forced an integer change in x.If we call the test change dx, we can find the difference between the new amount ($ \frac{1}{x+1} $) and the original ($ \frac{1}{x} $):

\[ \frac{1}{x + dx} - \frac{1}{x} = \frac{x}{x(x + dx)} - \frac{x + dx}{x(x + dx)} = \frac{-dx}{x(x + dx)} \]

After finding the total change (and its annoying algebra), we divide by dx to get the change on a “per dx” basis:

\[ \frac{-1}{x(x + dx)} \]

Now we remove the leftover dx, the measurement artifact:

\[ \frac{-1}{x(x + 0)} = - \frac{1}{x^2} \]

Phew!我们已经发现了1/x分割是如何随着人数的增加而变化的。

Let’s try it out: You are splitting a \$1000 bill among 5 people.A sixth person enters, how much money do you save?

You’ll personally save 1/5$ \cdot $1/6 = 1/30 of the total cost (cut your share into 6 pieces, give the new guy one portion to pay).That’s about 3%, or 30.Not bad for a quick calculation!

Let’s work it backwards: how large is our group when we’re saving about \$100 per person?Well, \$100 is 1/10 of the total.Since$ \frac{1}{3^2} \sim \frac{1}{10} $, we’ll hit that savings rate around x=3 people.

And yep, going from 3 to 4 people means each person’s share goes from \$333.33 to \$250 – about \$100.不坏!(如果我们将人数稍微增加一点，我们就能精确地达到这个数字。)

12.4Questions

We didn’t explicitly talk about scaling by a constant, such as finding the derivative of$ f(x) = 3x $.你能用乘积法则算出它是怎么变化的吗?(Hint: imagine a rectangle with a fixed 3 for one side, and$ x $for the other).

Now, how about the addition rule?How would$ f(x) = x + x + x $behave?

Next →Lesson 13: Finding Patterns In The Rules