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Differentiation

4.1 The Derivative and the Tangent Line Problem

We've already looked at the Tangent Line problem (Chapter 3), so now it's time to finish what was begun—to use limits to squash that secant line into a tangent line!

A brief recap first: the slope of a curve can make sense, but it changes from point to point—so we must talk about the tangent at a point; from a point, we can choose another point, and form a secant line through these two points; the issue is, what happens to the equation of the secant line as the second point approaches the first point?

Let's find out!

The Tangent Line Problem

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 $Δ x$ units away from the first point. Thus, the second point is at ( $x + Δ x$ , $f (x + Δ x)$ ).

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 $Δ x$ 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).

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

4.2 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 $\frac{d}{d x} [f (x) - g (x)] = \frac{d}{d x} [f (x)] - \frac{d}{d x} [g (x)]$ .

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 $y = x^{3}$ .

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

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

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

[6] Find $p^{'} (x)$ if $p (x) = 4 x^{2} + 2 x - 5$ .

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

4.3 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

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

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

$\frac{d}{d x} [3 x^{5} + 7 x^{3} - 12 x + 15] = 15 x^{4} + 21 x^{2} - 12$ ; $\frac{d^{2} y}{d y^{2}} = \frac{d}{d x} [15 x^{4} + 21 x^{2} - 12] = 60 x^{3} + 42 x$ ; $\frac{d^{3} y}{d y^{3}} = \frac{d}{d x} [60 x^{3} + 42 x] = 180 x^{2} + 42$ ; $\frac{d^{4} y}{d y^{4}} = \frac{d}{d x} [180 x^{2} + 42] = 360 x$ .

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

4.5 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). x² + 2xy + y² = 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 2xy must be handled with the product rule!

$2 x \frac{d x}{d x} + 2 [x \frac{d y}{d x} + \frac{d x}{d x} \cdot y] + 2 y \frac{d y}{d x} = 0$ ; 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!

An Image

[12] Find $\frac{d y}{d x}$ for 2x³ - 3y² = 7.

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

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