]> IB HL 2 - Chapter 02

Differentiation

2.1: The Derivative and the Tangent Line Problem

The Tangent Line Problem

We know all about lines—most importantly, we know all about the slope of a line. But why should the idea of slope be restricted to lines? Can't we say the graph below is steep…

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…while the next graph is rather shallow?

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The first graph was steep; it had a large (far from zero) slope. The second graph was shallow; it had a small (close to zero) slope.

The problem should be fairly evident—how do you describe the slope of this graph?

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In some places, it is steep—but in others, it's shallow!

For lines, slope is constant—it doesn't matter where you measure; you'll get the same slope.

For curves, the slope changes as you look at different points along the curve. There's the rub: our formula for slope uses any two points—we can't do that for curves.

We can approximate the slope of the curve (at a particular value of x) with a secant line. Observe: (the curve is blue; the secant line is red)

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First, the two points are (fairly) far apart—the red line (the secant line) doesn't show the slope of the curve. However, as the points get closer together…

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The secant line looks more and more like a tangent line, which shows the slope of the curve. The trick is to use limits to determine what happens to the secant lines as those points get closer together.

Let's define a few things: let f(x) be the curve in question, and let (x, y) be the point at which we want to find the equation of the tangent line. Note that we could (and should) label this point as (x, f(x)).

There are a number of ways to define the second point (that defines the secant line)—let's choose a method that avoids subscripts. In particular, let the second point be Δx units away from the first point. Thus, the second point is at (x + Δx, f(x + Δx)).

Now to the equation of the secant line—perhaps we should first find the slope?

$slope = \frac{change in y}{change in x} = \frac{f (x + Δ x) - f (x)}{(x + Δ x) - x} = \frac{f (x + Δ x) - f (x)}{Δ x}$

So here's the real problem: what happens to that slope as the second point gets closer to the first? What happens to this equation when Δx gets close to zero?

We've already tackled that—we should use limits!

$slope = lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}$

Of course, this is as far as we can go without actually knowing what f(x) is…

The Derivative

…but that's far enough! That equation gives us the first derivative of f(x).

$f^{'} (x) = lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}$

If this limit exists, then we can find the equation of the slope of the curve at any point—this can then be applied at the point in question to find a particular slope (and, ultimately, a particular equation of a tangent line).

There's one catch: the limit must exist…of course, we know that the limit exists if the left- and right-hand limits exist. So: f ′(x) exists if $lim_{Δ x \to 0^{-}} \frac{f (x + Δ x) - f (x)}{Δ x} = lim_{Δ x \to 0^{+}} \frac{f (x + Δ x) - f (x)}{Δ x}$ . This will be true at all values of x within the domain of polynomial and rational functions. It will not be true where the function has a sharp turn (a cusp; e.g., y = |x|), or a vertical tangent (e.g., $y = \sqrt[3]{x}$ ).

Differentiability and Continuity

Differentiable functions are necessarily continuous. Continuous functions need not be differentiable.

Notation

There are several notations for the first derivative of the function h(x): h′(x), $\frac{d y}{d x}$ , $\frac{d}{d x} [h (x)]$ …

Examples

[1.] Find the slope of $f (x) = 5 x + 7$ at the point (2, 17).

OK, I know—it's a line! We don't need to take the derivative! But let's do it anyway.

$lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x} = lim_{Δ x \to 0} \frac{[5 (x + Δ x) + 7] - [5 x + 7]}{Δ x} = lim_{Δ x \to 0} \frac{[5 x + 5 Δ x + 7] - [5 x + 7]}{Δ x} = lim_{Δ x \to 0} \frac{5 Δ x}{Δ x} = lim_{Δ x \to 0} (5) = 5$ . f′(x) = 5. Thus, the slope does not depend on the point—the slope if 5 at every point on this equation (duh!).

[2.] Find the slope of $g (x) = x^{2} + 1$ at the point (3, 10).

$lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x} = lim_{Δ x \to 0} \frac{[{(x + Δ x)}^{2} + 1] - [x^{2} + 1]}{Δ x}$ = $lim_{Δ x \to 0} \frac{[x^{2} + 2 \cdot x \cdot Δ x + {(Δ x)}^{2} + 1] - [x^{2} + 1]}{Δ x} = lim_{Δ x \to 0} \frac{2 \cdot x \cdot Δ x + {(Δ x)}^{2}}{Δ x} = lim_{Δ x \to 0} (2 x + Δ x) = 2 x$ . $g^{'} (x) = 2 x$ , so $g^{'} = 2 (3) = 6$ . The slope of g(x) at the point (3, 10) is 6—thus, the equation of the tangent line is $y - 10 = 6 (x - 3)$ .

2.2: Basic Differentiation Rules and Rates of Change

Rules

The Constant Rule

If f(x) is a constant function, then f(x) = c (where c is some real number). Let's look at the derivative in this case…

$f^{'} (x) = lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x} = lim_{Δ x \to 0} \frac{c - c}{Δ x} = lim_{Δ x \to 0} (0) = 0$

So for any constant function, f ′(x) = 0.

The Power Rule

Now let f(x) be a simple power function— $f (x) = x^{n}$ , where n is a rational number. Through careful use of the Binomial Expansion of Polynomials, you can prove that $f^{'} (x) = n \cdot x^{n - 1}$ .

The Constant Multiple Rule

Factoring allows us to show that $\frac{d}{d x} [c \cdot f (x)] = c \cdot \frac{d}{d x} [f (x)]$ .

The Sum and Difference Rules

Grouping (the associative property) allows us to show that $\frac{d}{d x} [f (x) + g (x)] = \frac{d}{d x} [f (x)] + \frac{d}{d x} [g (x)]$ and $\frac{d}{d x} [f (x) - g (x)] = \frac{d}{d x} [f (x)] - \frac{d}{d x} [g (x)]$ .

Trigonometric Functions

Functions other than polynomials require more work (tricks?) to use the definition of the derivative—for example, sine and cosine. Note the work on page 107 of the text—I'll not repeat it here.

Know that $\frac{d}{d x} [sin x] = cos x$ and $\frac{d}{d x} [cos x] = - sin x$ .

Rates of Change

The slope determines the amount of change in y for a unit change in x—thus, the slope is the rate of change for y.

So what, you say? Well, if the function determines the position of an object (y) for a given time (x), then the slope (the first derivative) is the change in position per unit change in time. Wait, that sounds familiar…change in position per change in time…that's velocity! And the change in velocity per change in time is acceleration!

Examples

[3.] Find n′(x) if n(x) = 7.

Easy! The derivative of a constant function is zero. n′(x) = 0.

[4.] Find $\frac{d y}{d x}$ if y = x³.

$\frac{d y}{d x} = \frac{d}{d x} [x^{3}] = 3 x^{2}$ .

[5.] Find $\frac{d}{d x} [5 x^{4}]$ .

$4 \cdot 5 x^{3} = 20 x^{3}$ .

[6.] Find p′(x) if p(x) = 4x² + 2x - 5.

$\frac{d}{d x} [4 x^{2} + 2 x - 5] = \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [2 x] - \frac{d}{d x} [5] = 8 x + 2 - 0$ .

2.3: The Product and Quotient Rules and Higher-Order Derivatives

The Product Rule

$\frac{d}{d x} [f (x) \cdot g (x)] = f^{'} (x) \cdot g (x) + f (x) \cdot g^{'} (x)$

The Quotient Rule

$\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{f^{'} (x) \cdot g (x) - f (x) \cdot g^{'} (x)}{{[g (x)]}^{2}}$

A result of the quotient rule is that we can now obtain derivatives of the other trigonometric functions:

$\frac{d}{d x} [tan x] = {sec}^{2} x$

$\frac{d}{d x} [cot x] = - {csc}^{2} x$

$\frac{d}{d x} [sec x] = sec x \cdot tan x$

$\frac{d}{d x} [csc x] = - csc x \cdot cot x$

Higher-Order Derivatives

The derivative of a function is another function. Naturally, you can take the derivative of that function, which results in the second derivative!

Second Derivative: $f^{″} (x) = \frac{d}{d x} [\frac{d y}{d x}] = \frac{d^{2} y}{{d x}^{2}}$

This can continue for quite some time…

Third Derivative: $f^{'''} (x) = \frac{d}{d x} [\frac{d^{2} y}{{d x}^{2}}] = \frac{d^{3} y}{{d x}^{3}}$

The n ^th Derivative: $f^{(n)} (x) = \frac{d^{n} y}{{d x}^{n}}$

Examples

[7.] Find $\frac{d}{d x} [x \sqrt{x}]$

$\frac{d}{d x} [x \sqrt{x}] = \frac{d}{d x} [x] \cdot \sqrt{x} + x \cdot \frac{d}{d x} [\sqrt{x}] = (1) \sqrt{x} + x (\frac{1}{2} x^{- \frac{1}{2}}) = \sqrt{x} + \frac{x}{2 x \sqrt{x}}$

[8.] Find j′(x) if $j (x) = \frac{2 x}{x^{2} + 1}$

$j^{'} (x) = \frac{(x^{2} + 1) (\frac{d}{d x} [2 x]) - (\frac{d}{d x} [x^{2} + 1]) (2 x)}{{(x^{2} + 1)}^{2}} = \frac{(x^{2} + 1) (2) - (2 x) (2 x)}{x^{4} + 2 x^{2} + 1} = \frac{2 x^{2} + 2 - 4 x^{2}}{x^{4} + 2 x^{2} + 1} = \frac{2 - 2 x^{2}}{x^{4} + 2 x^{2} + 1}$

[9.] Find $\frac{d^{4} y}{{d x}^{4}}$ if $y = 3 x^{5} + 7 x^{3} - 12 x + 15$ .

$\frac{d}{d x} [3 x^{5} + 7 x^{3} - 12 x + 15] = 15 x^{4} + 21 x^{2} - 12$

$\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} [15 x^{4} + 21 x^{2} - 12] = 60 x^{3} + 42 x$

$\frac{d^{3} y}{{d x}^{3}} = \frac{d}{d x} [60 x^{3} + 42 x] = 180 x^{2} + 42$

$\frac{d^{4} y}{{d x}^{4}} = \frac{d}{d x} [180 x^{2} + 42] = 360 x$

2.4: The Chain Rule

The Chain Rule

We've looked at lots of ways to combine functions—sums, differences, products, quotients…that leaves only one other way!

Compositions! Enter the Chain Rule:

$\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$

The General Power Rule

If you combine the Power Rule and the Chain Rule, you get the General Power Rule:

$\frac{d}{d x} {[f (x)]}^{n} = n \cdot {[f (x)]}^{n - 1} \cdot f^{'} (x)$

Simplifying Derivatives

Just do it!

Examples

[10.] Find the first derivative of $v (x) = \sqrt{3 x^{2} + 9 x - 5}$

We can decompose $v (x)$ into $f (x) = \sqrt{x}$ and $g (x) = 3 x^{2} + 9 x - 5$ , so that $v (x) = f (g (x))$ .

Then $v^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = \frac{1}{2} {[g (x)]}^{- \frac{1}{2}} \cdot (6 x + 9) = \frac{1}{2} {(3 x^{2} + 9 x - 5)}^{- \frac{1}{2}} (6 x + 9) = \frac{6 x + 9}{2 \sqrt{3 x^{2} + 9 x - 5}}$

2.5: Implicit Differentiation

Explicit Equations

Almost all of the equations that you've looked at have been in explicit form—in other words, the function has been solved for one variable (typically y, which is what makes it a function). Taking derivatives of explicit functions has been covered quite thoroughly, I think…

Implicit Equations

An implicit equation is one that has not been solved for a variable—perhaps because it is difficult to do so; perhaps because it is impossible. Maybe the equation is a function; maybe it's only a relation.

The equation xy = 1 is in explicit form, and can be easily solved for y (it is a function). $x^{2} + 2 x y + y^{2} = 1$ is in implicit form, and cannot easily be switched to explicit since it does not represent a function.

Implicit Differentiation

Naturally, you can find the slope of any curve—function or not. When the equation is in implicit form, you must use implicit differentiation. There are two tricks to doing this: [1] remembering the chain rule, and [2] factoring/solving for $\frac{d y}{d x}$ . Perhaps some examples would be the best way to show you how to do this…

Examples

[11.] Let's find $\frac{d y}{d x}$ for the equation x² + 2xy + y² = 1.

TIP: think of x and y as functions! Thus, the derivative of x² is $2 x \frac{d x}{d x}$ , and the derivative of y² is $2 y \frac{d y}{d x}$ . Also, the derivative of $2 x y$ must be handled with the product rule!

$2 x \frac{d x}{d x} + 2 [x \frac{d y}{d x} + \frac{d x}{d x} y] + 2 y \frac{d y}{d x} = 0$ ; or, since $\frac{d x}{d x}$ is just 1, $2 x + 2 [x \frac{d y}{d x} + y] + 2 y \frac{d y}{d x} = 0$

Simplify. $2 x + 2 x \frac{d y}{d x} + 2 y + 2 y \frac{d y}{d x} = 0$ .

Now we need to solve for $\frac{d y}{d x}$ —group and factor! $2 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = - 2 x - 2 y$ ; $\frac{d y}{d x} [2 x + 2 y] = - 2 x - 2 y$ ; $\frac{d y}{d x} = \frac{- 2 x - 2 y}{2 x + 2 y} = \frac{- 2 (x + y)}{2 (x + y)} = - 1$ .

Since the derivative is a constant, this should tell you that this implicit equation is a line!

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[12.] Find $\frac{d y}{d x}$ for $2 x^{3} - 3 y^{2} = 7$ .

$6 x \frac{d x}{d x} - 6 y \frac{d y}{d x} = 0$ ; $6 x - 6 y \frac{d y}{d x} = 0$ ; $6 x = 6 y \frac{d y}{d x}$ ; $\frac{d y}{d x} = \frac{6 x}{6 y} = \frac{x}{y}$ .

2.6: Related Rates

We know all about relating two variables in equations—what we haven't done is investigate how the rate of change of one variable is related to the rate of change of another variable (or several other variables), with respect to a another variable (time).

Did I mention that we're talking about word problems—applications of calculus?

The Technique

You're going to need to write an equation that relates the two variables. Then, take the implicit derivative with respect to t (time). Now you should be able to plug in the given values and solve for the requested item.

Perhaps this should be illustrated with some examples…

Examples

[13.] A rock is dropped into a pond, which causes a circular ripple. The ripple moves away from the center (where the rock was dropped) at 1m/s. How fast is the area of the circle changing when the radius is 3m?

Let's inventory what was given in the problem.

The ripple moves away from the center at 1m/s: this is the rate of change of the radius; $\frac{d r}{d t}$ .

How fast is the area of the circle changing…: this is asking for the rate of change in the area; $\frac{d A}{d t}$ .

… when the radius is 3m: r = 3.

So, the variables are radius, area, and time. We need to write an equation that relates the area and the radius. Any ideas?

Of course! $A = π r^{2}$ . Now that we've got an equation, we need to differentiate with respect to t.

$\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ .

Now, we're trying to solve for $\frac{d A}{d t}$ : that means we need to plug in value for everything else—r and $\frac{d r}{d t}$ .

Joy! They were given as $\frac{d r}{d t} = 1$ and r = 3!

$\frac{d A}{d t} = 2 π (3) (1) = 6 π$ .

When the radius is 3m, the area is changing at 6πm²/s.

[14.] Gravel is being poured out of a container onto the ground at a rate of 2yd³/min. The gravel forms a cone as it lands, where the height of the cone is about half of the diameter. How fast is the height of the cone rising when the diameter of the pile is 3yd?

$\frac{d V}{d t} = 2$ , r = 1.5, h = ½d = r, and we're asked about $\frac{d h}{d t}$ . We need an equation that relates V, r and h…

$V = \frac{1}{3} π r^{2} h$ , so $\frac{d V}{d t} = \frac{π}{3} [2 r \frac{d r}{d t} h + r^{2} \frac{d h}{d t}]$ . Plug in! $2 = \frac{π}{3} [2 (1.5) \frac{d r}{d t} h + {(1.5)}^{2} \frac{d h}{d t}]$ .

Oops—what are we going to do about $\frac{d r}{d t}$ and h? Well, h = r, so h = 1.5. What about $\frac{d r}{d t}$ ? Well, if h = r, then $\frac{d h}{d t} = \frac{d r}{d t}$ ! So we get $2 = \frac{π}{3} [2 (1.5) \frac{d h}{d t} (1.5) + {(1.5)}^{2} \frac{d h}{d t}]$ . Now we just need to solve for $\frac{d h}{d t}$ …

Factor! $2 = \frac{π}{3} [2 (1.5) (1.5) + {(1.5)}^{2}] \frac{d h}{d t}$ means $\frac{6}{π [2 (1.5) (1.5) + {(1.5)}^{2}]} = \frac{d h}{d t}$ , $\frac{6}{π [2 {(1.5)}^{2} + {(1.5)}^{2}]} = \frac{d h}{d t}$ , $\frac{6}{3 π {(1.5)}^{2}} = \frac{d h}{d t}$ , $\frac{2}{π {(1.5)}^{2}} = \frac{d h}{d t}$ . That gives an approximate rate of change of 0.2829yd/min.

[15.] Stranded on Tatooine, R2-D2 and C-3PO have decided to split up in the desert. R2-D2 moves due north at 7km/hr, and C-3PO moves due east at 5km/hr. How fast is the distance between them increasing after 2 hours?

Let's let n = distance north (of R2-D2), e = distance east (of C-3PO), and D = distance between the two. So we're given $\frac{d n}{d t} = 7$ , $\frac{d e}{d t} = 5$ , and t = 2.

The only equation we can write that relates these is $n^{2} + e^{2} = D^{2}$ . Of course, that means $2 n \frac{d n}{d t} + 2 e \frac{d e}{d t} = 2 D \frac{d D}{d t}$ . What are we going to plug in for n, e and D? How will we use the value t = 2?

Of course! At t = 2, n = 14, e = 10, and $D = \sqrt{14^{2} + 10^{2}} = \sqrt{196 + 100} = \sqrt{296}$ .

$2 (14) (7) + 2 (10) (5) = 2 \sqrt{296} \frac{d D}{d t}$ , so $\frac{d D}{d t} = \frac{296}{2 \sqrt{296}} = \frac{148}{\sqrt{296}}$ , or approximately 8.602km/hr.

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