Differentiation

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Limits and Derivatives

How could we find the slope of a parabola? Unlike a line, the steepness of a parabola varies, so we need to interpret the steepness of the curve at a point. For instance, look at x=2 on the graph of f(x)=x2 below.

What can we tell about the slope at that point? What if we utilize our knowledge of the slope of a straight line to estimate the slope at x=2 by connecting it to a nearby point on the graph of f.

Here, we have come up with an estimate for the slope, but how can we improve our estimate?

Discussion

What if, as shown above, we use (3,9) instead of (4,16) for our slope approximation? Explain whether our estimate improves.
Use the graph below to estimate the slope at x=2 as closely as possible.
Powered by Desmos
State an x-value that would improve your approximation from part (b).
Is there an x-value that would give a "best," or perfectly accurate, approximation of the slope?

Part (a)

Visually, it looks like the line connecting (2,4) and (3,9) clings closer to the parabola than the one connecting (2,4) and (4,16). Thus, the line connecting (2,4) and (3,9) seems to give a better approximation. This makes sense because we know that the slope of the parabola is changing (getting steeper as x goes to the right). Furthermore, since we want the slope exactly at x=2, we expect the estimate to get better the closer the other point gets to x=2.

Part (b)

As we bring the other point nearer to x=2 on the graph, we see that the slope approximation gets closer and closer to 4. We can get as close as x=2.01 to x=2, where the slope of the line passing through our two points is 4.01:

Part (c)

To get even closer to the true slope at x=2, we simply choose our second point still nearer to 2 than 2.01. For example, take

c=2.001.

Then

f(2.001)=(2.001)2=4.004001

and the slope is

2.001−2f(2.001)−f(2)=0.0014.004001−4=4.001,

which is a better approximation to the true slope 4 than our previous 4.01. You could go even closer—say c=2.0001.

Part (d)

Imagine you pick any other point on the parabola—say at x=c=2—and draw the straight segment linking (2,4) and (c,c2). You measure its steepness by “rise over run,” that is, how much you go up divided by how far you go across.

No matter which c you choose apart from 2, that ratio always misses the true steepness of 4 by a bit. If you try to force the ratio to land exactly on 4, the only way to do it algebraically is to collapse your horizontal gap to zero—meaning you’d have to let c actually equal 2. But then your two points coincide, so there’s no “run” to measure at all, and no straight‐line slope you can talk about.

In plain terms:

– Any second point strictly left or right of x=2 gives a rise/run different from 4.

– The only way to nail 4 exactly is to let the two points merge, destroying the very idea of a line between them.

So there is no distinct x-value that makes your two-point estimation perfectly match the curve’s steepness at 2. The best you can do is pick points ever closer to 2: as c inches in on 2, the rise/run number creeps closer and closer to 4, approaching perfection but never hitting it.

We just explored the concept of a limit. The basic idea of a limit is that as x approaches a given value a, f(x) approaches a value too. In our exploration, we saw what happened to slope (x2−x1y2−y1=x2−2y2−4) as x2 approaches 2.

When we think about limits as x→a, we consider x getting infinitesimally close to a but never reaching it. By infinitesimal, we mean infinitely close.

Basic concept of a limit

The limit x→alimf(x) is the value f(x) approaches as x approaches a.

Types of limits:

Trivial limit: a limit that is simply equivalent to the value of the function evaluated at a.

x→3lim(x2+2)=32+2=11

Rational limit: a limit of a rational function as x approaches a value at which the function is undefined.

x→3lim(x−3x(x−3))=x→3lim(x)=3x→3lim((x−3)(x−1)x2)=∞

Infinite limit: a limit as ∣x∣ gets arbitrarily large. If a function grows without bound as ∣x∣ gets larger, the limit evaluates to ∞ (or −∞). If the function has an asymptote y=k, the limit is k. Infinite limits are undefined for oscillating functions since they do not approach a single value.

x→∞lim(2x−13x+2)=x→∞lim(2−x13+x2)=23x→−∞lim(3x2−6x+1)=∞x→∞limsinxdoes not exist

Slope as a Limit

The IB may test your understanding of the gradient of the curve as the limit of

m=x2−x1y2−y1

as (x2−x1) goes to zero.

Checkpoint

Calculate x→−∞limex.

Select the correct option

Each type of limit may be interpreted graphically:

Trivial limits are simply the value f(a).

Rational limits will appear as vertical asymptotes or holes on a graph.

Infinite limits may be evaluated as a number if there is a horizontal asymptote. Otherwise, if a curve increases or decreases without bound, these limits evaluate as ∞ or −∞.

Limit from a graph

x→1limf(x)=2

x→∞limf(x)=23

x→−∞limf(x)=23

One way to conceptualize a limit is that in the graph above, we can get as close to an output of 2 as we want near x=1.

Focus on the right side of the curve as it approaches the hole, we could find an x so that 1.9<f(x)<2, 1.99<f(x)<2, or even 1.9999<f(x)<2.

The key idea here is that since x→1limf(x)=2, pick a value as close to 2 as you want, and we can find an x-value close enough to 1 so that f(x) is even closer to 2:

Exercise

Part of the graph of y=f(x) is shown in the diagram below.

Find

x→1limf(x),
x→∞limf(x).

Select the correct option

Discussion

Values of a function f are shown at values of x surrounding 2.

x	1.9	1.99	1.999	1.9999	2.0001	2.001	2.01	2.1
f(x)	−7.053	−0.753	−0.0754	−0.0075	0.0075	0.0754	0.753	7.053

From the table, find x→2limf(x).

As the table shows, whenever x is closer to 2 (whether just below or just above), the corresponding values f(x) get closer to 0. For example:

x=1.9x=1.99x=1.999x=1.9999⟹f(x)=−7.053,⟹f(x)=−0.753,⟹f(x)=−0.0754,⟹f(x)=−0.0075,x=2.0001x=2.001x=2.01x=2.1⟹f(x)=0.0075,⟹f(x)=0.0754,⟹f(x)=0.753,⟹f(x)=7.053

In each case, as x moves nearer to 2, f(x) moves nearer to 0. Hence

x→2limf(x)=0

Limit from a table

Given a table of values:

xf(x)0.91.620.991.91210.9991.9972201……

x→1limf(x)=2

Exercise

The following table shows the value of f(x) for certain values of x

x	2.9	2.99	2.999
f(x)	8.41	8.9401	8.994001

Using the table above, estimate the value of x→3limf(x).

Select the correct option

Now that we have explored the limit, we can use it to improve our approximation of the slope to an exact value.

We start with by zooming in even more on the graph of f(x)=x2 with a line connecting the points on f at x=2 and x=2. You can now drag the point even closer to x=2, and watch the slope get even closer to 4.

As you can see, we can get as close as 4.0001 to 4 by moving the x-value to 2.0001. Now, consider that as close as 2.0001 is to 2, 0.0001 is a finite distance between points, and with limits, we may evaluate what happens as x gets infinitesimally close to 2 with

x→2lim2−x4−f(x) =x→2lim2−x4−x2=x→2lim2−x(2−x)(2+x)=x→2lim(2+x)=4.

This is our first exploration of the derivative, which returns the instantaneous slope of a function by taking the limit of the slope of the line connecting a point on f to points that are infinitesimally nearby.

Discussion

The following equation defines the derivative for the entire function f:

f′(x)=h→0limhf(x+h)−f(x).

Show that the new equation for the derivative is also a slope between two infinitesimally close points.
Using the new definition of the derivative, find f′(x) given that f(x)=x2.

Part (a)

The definition

f′(x)=h→0limhf(x+h)−f(x)

is another way to express the slope between two points on the graph of f that are very close together.

Recall that the slope between two points (x1,y1) and (x2,y2) is given by

m=x2−x1y2−y1

If we let x1=x and x2=x+h, then y1=f(x) and y2=f(x+h). Substituting into the slope formula gives

m=(x+h)−xf(x+h)−f(x)=hf(x+h)−f(x)

So, the expression inside the limit in the definition of the derivative is just the slope between the points (x,f(x)) and (x+h,f(x+h)).

Taking the limit as h→0 means we are finding the slope between two points that are infinitesimally close together. Therefore, the definition

f′(x)=h→0limhf(x+h)−f(x)

still represents the slope between two infinitesimally close points on the curve.

Now, it is reasonable to conclude that even though the two equation for the derivative look disimilar, they are indeed equivalent since they both give the slope of a curve "at a point."

Part (b)

Using the definition

f′(x)=h→0limhf(x+h)−f(x)

with f(x)=x2, write

f′(x)=h→0limh(x+h)2−x2

Expand in the numerator:

(x+h)2−x2=(x2+2xh+h2)−x2=2xh+h2

f′(x)=h→0limh2xh+h2=h→0limhh(2x+h)=h→0lim(2x+h)=2x

Answer: 2x

Hence, if we plug in 2, we get the same answer - 4 - as we did graphically!

Limit definition of derivative

The derivative of f(x) is denoted f′(x) and is given by

f′(x)=h→0limhf(x+h)−f(x)📖

Gradient

For a curve y=f(x), f′(x) is the gradient or slope.

Exercise

Given that f′(x)=x2−x, find the slope of the graph of y=f(x) when x=3.

Select the correct option

Graphing a derivative with a GDC

You can graph f′(x) using the following steps:

Press the Y= key.
In one of the available function lines (e.g. Y_1), enter the expression for f(x).
In another available line (e.g. Y_2), input the derivative function usingMATH then 8:nDeriv( in the following format:
dXd(Y1(x))∣X=X
To enter Y1, press VARS then scroll to Y-VARS and select FUNCTION then Y1.
Press GRAPH to display both the original graph f and the derivative f′.
The graph of f′ may take a little bit longer depending on the original function.

After graphing f′, you may use all the other graphing functions on the calculator (intersect, zero, and value).

Discussion

Consider the area of an equilateral triangle A as a function of side length s: A(s)=4√3s2. Calculate A′(s).
Describe what the value of A′(s) tells us about the relationship between area and side length of an equilateral triangle.

Part (a)

We have

A(s)=4√3s2

dsdA=4√3⋅2s=2√3s

Hence

A′(s)=2√3s.

Part (b)

From part (a) we have

A′(s)=2√3s

Recall that the derivative A′(s) is exactly the slope of the curve A(s) at the point s. In other words, it is the instantaneous rate of change of the area with respect to the side length.

Because for all s>0 we have

A′(s)=2√3s>0

the graph of A(s) is rising: increasing s always increases A. Moreover, since A′(s) itself grows linearly with s, the slope becomes steeper as the triangle gets larger, which makes sense since area is proportional to side length squared.

Concretely, a very small increase ds in the side gives an increase dA, given approximately by

dA≈A′(s)ds=2√3sds

so for a larger s each extra unit of side adds more area than it would for a smaller triangle.

Since derivatives describe the rate of change of one variable with respect to another, we introduce another notation for derivatives: A′(s)=dsdA, which emphasizes that the derivative shows how A will respond to a small change in s.

Rate of Change

dxdy is the rate of change of y with respect to x. That is, dxdy tells us how much y changes in response to a change in x.

If y=f(x), then dxdy=f′(x).

You may see other variables than y and x with this notation. Note that d□d□ will always give the rate of change of the top with respect to the bottom. Hence, if you see dydx, this gives how much x changes in response to a change in y, which is not the slope of a curve y=f(x).

Derivative of xⁿ where n is an integer

f(x)=xn, n∈Z⇒f′(x)=nxn−1📖

Checkpoint

Find the derivative of x5.

Select the correct option

Derivatives of sums and scalar multiples

dxd(af(x))=af′(x)🚫

dxd(f(x)+g(x))=f′(x)+g′(x)🚫

dxd(af(x)+bg(x))=af′(x)+bg′(x)🚫

Checkpoint

Find the derivative of 2x3+4x2−6x+12.

Select the correct option

Exercise

Find dxd(10x3−x46).

Select the correct option

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