Sec 3

Section One | Derivatives A derivative is defined by Merriam-Webster as "the limit of the ratio of the change in a function to the corresponding change in its independent variable as the latter change approaches zero". This is rather hard to think about in those words, however, so instead, a derivative is more simply defined as the slope at any point on a function. This slope at that point is defined as the slope of the tangent line. There are many ways to compute a derivative, ranging from the simple to the complex, but we'll get to those later. Instead, let's return to the definition that Merriam-Webster gives us. To put this definition in equation form, you would get the following. Note, h = ∆ x. limh →0fx+h- f(x)h So, as the change in x gets smaller, the average rate of change will eventually become the instantaneous rate of change. This equation can also be interpreted as the change in f(x) over the change in x. You can use this formula for calculating a derivative be substituting the function you want to find the derivative for into the equation for x. So, let’s say you want to find the derivative of x3. So, plug in x3 for the x’s in the above equation, and you will come up with this result: = (x+h)3- x3h = x3 + 3x2h + 3x2h2+ h3- x3 h = h(3x2+ 3xh+ h2)h = 3x2 So, the derivative of x3 is 3x2. This can be applied to many functions, however. For simple functions, the way to find a derivative is to bring the exponent that x is raised to down as a coefficient, and then subtract the exponent by one to get the derivative. Below is a table of examples of this property: Function x3 12 19x4 (5x2 + 12x +19) x (or, x1/2)

Derivative 3x2 0 76x3 (10x + 12) 12x (or, ½x-1/2)

Similarly, to evaluate a derivative at a point (that is, give the numeric value for the slope of the tangent line at that point), you simply plug in the x of the point that you are given. So evaluating the previous table at x = 2 will give you a set of slopes. Function

Derivative

Derivative at x

3

x 12 19x4 2 (5x + 12x +19) x (or, x1/2)

2

3x 0 76x3 (10x + 12) 12x (or, ½x-1/2)

=2 12 none 608 32 122

EXAMPLE PROBLEMS:

Section Two | Continuity & Differentiability There are not always derivatives at every point of a function, however. To be differentiable at every point x on a function, the function must be continuous. A simple definition of "continuous" is, if drawing the function, you have to lift your pencil from the paper at any time to draw it, and it is not a continuous function. This means that functions with asymptotes, holes, jumps, and other irregularities. There are no derivatives at the ends of piecewise functions and at cusps as well.

There is also another special kind of function that has certain conditions for its differentiability. With functions with the form xfraction, they must meet two requirements to be differentiable at all points. To figure out those conditions, let's use an example: General Function xeven/odd

Example Function x2/5

Derivative 25x-35

Derivative at x = 0 Undefined

xodd/odd

x5/3

53x23

0

xodd/even

x1/2

12x-12

Undefined

With these general functions, the only form to have a derivative at all is xodd/odd. However, this is not always true either. For example, the function x3/5’s derivative is undefined at zero. Whether the derivative is zero or undefined depends on the following two criteria: 1. “X” is raised to a fraction that is an odd number over an odd number 2. This fraction is > 0 Cusps, which are made from the other fractional exponents, are not differentiable at all values of x either. To see why that is, look at this graph of a cusp: If you imagine a tangent line following this graph’s path, then at the point where the negative slopes become positive (the cusp point), a special tangent line is formed. This line is a horizontal line, which has an undefined slope. That is why this function isn’t always differentiable, yet it is, technically, continuous. On the other hand, when a derivative at a point has a value of zero, this tangent line is vertical, since that would mean its slope is zero.

Section Three | Tangent and Normal Lines Tangent lines are, as mentioned, the line that is tangent to the graph at some point x. To compute the equation for the tangent line, you need the slope of the tangent

line (given by evaluating a derivative) and the ordered pair of the point that this line was evaluated at. For tangent lines, using the point-slope form of an equation from a line is easiest. The point-slope equation of a line is m(x – x1) = y – y1, where m is the slope, and (x1, y1) represents the ordered pair. So, for examples of how to form a tangent line, let’s use those examples. However, I am leaving out f(x) = 12, since this function does not have a proper equation. Function

Derivative 3x2

Derivative at x =2 12

Ordered Pair (2, 8)

x3 19x4

76x3

608

(2, 304)

(5x2 + 12x +19) x (or, x1/2)

(10x + 12)

32

(2, 19)

12x (or, ½x1/2 )

122

(2, 2)

Tangent line 12(x – 2) = y – 8 608(x – 2) = y – 304 32( x – 2) = y – 19 122(x – 2) = y 2

A normal line is a line which is perpendicular to the tangent line to the curve at that point. This line can be calculated by taking the negative reciprocal of the slope. So, for the examples above, their normal lines are in the table below: Normal Line -112(x – 2) = y – 8 -1608(x – 2) = y – 304 -132( x – 2) = y – 19 -22(x – 2) = y - 2

EXAMPLE PROBLEMS:

Section Four| Local Linearization

Local linearization is the concept that derivatives can be used to calculate the value of a function. Linearization does not find the actual value of "a", but instead approximates the output of a function near x = a. The equation for the linearization of a function at this x = a is: y=fa+ f'(a)(x-a) This equation will give you another equation for approximating a value or a function. Let’s say you want to approximate 9.003. We can use the knowledge that 9=3 to help us. To linearize this value into a function, simply replace that value of 9.003 with x. fx= x (and at x=a…) y= a+ 12a(x-a) Then, use our known, close value of 9, so that a = 9. Then, after finding this equation, we want to approximate this to x = 9.003, so plug this in. y=3+ x-96 y=3+ 9.003-96=3.001 The actual value of 9.003 is close to 3.0004100, so the error in this approximation is about 2%.

EXAMPLE PROBLEMS:

Section Five | Rates of Change The general function of a derivative mentioned in section one is very closely tied to the idea of rates of change. After all, as that value h gets smaller and smaller, it eventually becomes the instantaneous rate of change rather than the average rate

of change that the overall function represents. However, geometrically, these ideas are very different. The average rate of change (A.R.C.) is represented by a line, just like the instantaneous rate of change (I.R.C.), but the A.R.C. is graphically represented by a secant line, while the I.R.C. is represented by a tangent line. The secant line connects two points, while the tangent line only touches a curve in one point. The A.R.C. is the slope of the secant line, while the I.R.C. is the slope of the tangent line. To demonstrate the average rate of change, a graph and an equation are shown below.

A.R.C.= f(x2) -f(x1)x2- x1

The average rate of change, in the application of driving, is the average total speed for a trip. The instantaneous rate of change, however, is the speed at an exact moment, and the one you read from your speedometer.

EXAMPLE PROBLEMS:

Section Six | Position, Velocity, and Acceleration Position, velocity, and acceleration are all related concepts when it comes to derivatives. After all, they are derivatives of each other. It’s possible to take the derivative of a function multiple times, and this function will tell you something new or different each time, especially in relation to motion. The original function represents the position of the function. The derivative of the function is the velocity.

The second derivative of the function is representative of acceleration. There are even further, though uncommon applications, such as a jerk, snap, crackle, and pop. To find these values, you simply take the derivative of the previous derivative. Of course, for most functions, there is a point where you cannot take the derivative any further, because it eventually becomes a number rather than just a function. Notable exceptions to this would be trigonometric functions. This property is especially useful because since you evaluate a derivative at a single point, this means that you would be finding the velocity and acceleration at a specific time rather than over an interval. Let’s take the function 4x4 + 12x3 - 17x2, for an example. So, when x = 1; Degree F(x) F’(x) F’’(x)

Which Means Position Velocity Acceleratio n

Derivative

At x = 1

n/a 16x3 + 35x2 34x 48x2 + 70x 34

-1 17 84

EXAMPLE PROBLEMS:

Section Seven | Approximate a Derivative Derivatives can also be approximated from graphs, or from tables of values.

From a graph, you just need to remember what a derivative is. Since a derivative is the slope of the original function, then you only need to generally keep track of the slope of the original function. For example, when the original graph forms a minimum or maximum, the derivative will be at the x-axis. Since in a minimum or maximum, the slope is going from negative to positive or from positive to negative, the graph of the slopes would have to cross the x-axis at this point. Similarly, just like you can use the graph of f(x) to approximate the derivative, you can use the graph of the derivative to approximate f(x). It’s simply applying those same concepts and relations, but the other way around. A graphic representation of this is below. The original function has a yellow background, while the derivative has a blue background.

To approximate a derivative from a table of values, you simply look at the values around them and subtract them. This is easier to see in an example, which you'll see below. We want to approximate the derivative at g(2x) and g'(0.2), given a table of values of the original function, f(x). x f(x)

0.1 3.17 1

0.2 3.27 9

0.3 3.70 5

0.4 4.07 6

0.5 4.66 6

0.6 4.80 4

Since we want 2x, the actual value we will be looking at on the table is 0.4, since 2(0.2) = 0.4. Since 0.3 and 0.5 are the values around 0.4, we will be subtracting the f(x) values and then dividing by the difference of the x values. So; g2x≈24.666-3.7050.5-0.3 which is 2.961.2 g2x≈9.61

EXAMPLE PROBLEMS:

Section Eight | Differentiation Formulas As I mentioned in the earlier sections, there are lots of different ways to find derivatives, and they can only be used in particular situations. The basic rule that we have been using thus far is known is simplified into this equation: y= xn →y'= nxn-1 When you have a function whose x-value is in an exponent, you instead use the following rule. In this case, a is a non-zero number. y= af(x) →y'= af(x) ×lna×f'(x) So, if y = 512x+4; y= 512x+4

y'=512x+4 ×ln5 ×12 EXAMPLE PROBLEM:

When there is a logarithm in the function, you use this rule: y=lnfx→y'= f'(x)f(x) So, if y = ln(3x2 + 2x) y = ln(3x2 + 2x) y'= 6x+23x2+ 2x EXAMPLE PROBLEM:

When there is a function inside another function, you use a rule called the Chain Rule. F(x) is some function, while g(x) is another. y=fgx→y'= f'gx×g'(x) So, if y = (x2 + 9)3; y=(x2+ 9)3 y'= 6x(x2+ 9)2 EXAMPLE PROBLEM:

When two functions are being multiplied together, you use the Product Rule. y=fxgx→y'=f'xgx + fxg'(x) So, if y = sec(x)tan(x); y=secxtanx y'=secxtan2x+ sec3(x) EXAMPLE PROBLEM:

Similarly, when two functions are being divided by each other, you use the Quotient Rule. y= f(x)g(x) →y'= f'xgx - fxg'(x)g(x)2 So, if y = 4x32x+7;

y= 4x32x+7 y'= 12x22x+7- 8x3(2x+7)2= 16x3+ 84x2(2x+7)2 EXAMPLE PROBLEM:

Trigonometric functions are often used as a combination of the above rules, but they have special derivatives of their own. These derivatives are summarized in the table below. Function y = sin x y = cos x y = tan x y = csc x y = sec x y = cot x

y' y' y' y' y' y'

Derivative = cos x = -sin x = sec2 x = -csc(x)cot(x) = sec(x)tan(x) = csc2 x

Similarly, inverse trigonometric functions have their own derivatives.

EXAMPLE PROBLEMS:

Function y = arcsin x

y' = 11- x2

Derivative

y = arccos x

y' = -11- x2

y = arctan x

y' = 1x2+ 1

y = arccsc x

y' = -1|x|x2- 1

y = arcsec x

y' = 1|x|x2- 1

y = arccot x

y' = -1x2+ 1

Section Nine | First and Second Derivative Tests Derivatives can also tell you about the behavior of the original function. The first and second derivative tests are used to discern specific characteristics about the original function that the derivative came from. The first derivative test is used to find the extrema of f(x) while the second derivative test is used to find inflection points. The extrema are the maximums and minimums of a function, while the inflection point is where concavity changes. To conduct the first derivative test: 1. Find f’(x) 2. Find points where f’(x) = 0, or where f’(x) is undefined 3. Investigate the sign changes of f using a sign chart a. If f’(x) changes from positive to negative, this is a maximum b. If f’(x) changes from negative to positive, this is a minimum c. If f’(x) does not change signs, this point is not an extreme To conduct the second derivative test: 1. Find f’’(x) 2. Find points where f’’(x) = 0, or where f’’(x) is undefined 3. Investigate the sign changes of f using a sign chart a. If f’’ > 0, f(x) is concave up b. If f’’ < 0, f(x) is concave down As an example of both of these functions, take the function f(x) = x3 – 5x2 - 8x. The first derivative of this function is 3x2 – 10x – 8, and the second derivative is 6x – 10. First Derivative Test F’(x) = 3x2 – 10x – 8 0 = 3x2 – 10x – 8 = (3x+2)(x-4) X = -23, 4

So, this function’s critical points are -23 and 4. So to make a sign chart, look at the values around the critical points to see if they are positive or negative.

According to this sign chart, x = -23 is a maximum, while x = 4 is a minimum. This is not an acceptable way to phrase and support this conclusion, however. Instead, you would say the following:

“f’ > 0 when x < -23

and f’ < 0 when -23 < x < 4, so x = -23 gives a maximum. f’

< 0 when -23 < x < 4, and f’ > 0 when x > 4, so x = 4 gives a minimum.” Second Derivative Test F’’(x) = 6x – 10 0 = 6x - 10 X = 53 So, this function’s critical point is 53 . Again, make a sign chart.

There is indeed an inflection point at x = 53. However, once again, that must be said a bit better. “F’’(x) > 0 when x > 53 and h’’ < 0 when x < 53 , so x = 53 gives an inflection point. EXAMPLE PROBLEMS:

Section Ten | Implicit and Explicit Differentiation Explicit differentiation is what we have been doing up until now. These are normal functions that are functions of x. However, implicit differentiation means that these functions are written as different functions of x. For example, the standard equation of a circle, x2 + y2 = 25, is an implicit function, since y is on the same side as x. It's possible to write this implicit function explicitly, but to find a derivative, there is no need to. In addition, there are some functions that cannot be written explicitly. In these cases, you just have to treat y, or whatever other variable is representing y, as another part of the equation you're deriving. However, since this variable has a "value" associated with it, what you are doing needs to be specifically stated. So, for x2 + y2 = 25; x2+ y2=25 2x+2yy'= 0 Then you simply need to get y’ on its own. 2yy'= -2x y'= -2x2y= -xy EXAMPLE PROBLEMS:

Section Eleven | Related Rates Related rate problems are an application of implicit differentiation. The defining quality of related rates problems is that there are generally two unknowns, but these unknowns are related in a way that you can use them in a differential equation, or combine two known formulas defined by another variable. Let's demonstrate this with an example: Air is being pumped into a spherical balloon at a rate of 4 cm3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 16 cm. The first thing to notice here are the two equations that are related: volume (V(t)) and radius (r(t)). The rate at which the volume is increasing is known, 4 cm3/min. Since rates of chance are analogous to a derivative, you know that V'(t) = 4. We want to know the rate at which the radius is changing. So, thinking of that unknown rate as a derivative; r't= ?| rt= d2=8 Now we just relate these two things to each other. Using the equation for the volume of the sphere, we can do just that. Vt= 43πr(t)3 We will need to use implicit differentiation on the above formula to get a useable form, but then we only need to plug in the values we already know. V'= 4πr2r' 4= 4π(8)2r' r'= 164π cm/min EXAMPLE PROBLEMS: