1INTRODUCTION

Although algebra, trigonometry, and geometry are of fundamental importance to the mathematician and physicist, a wide variety of technical problems cannot be solved using only these tools of mathematics. Many problems must be solved using methods of calculus. Calculus comprises of 2 fields, Differential Calculus (Differentiation) and Integral Calculus (Integration). This is a quick review of some differentiation concepts. It is meant to cover the basics which you will need for the Introductory Physics Course. It is not meant to teach you everything there is to know about Differentiation, but should give you a solid base for learning more on your own.

DIFFERENTIATION Getting Used to the Terminology Differentiation A process in mathematics which provides a derivative Derivative Something that describes the rate of change of one quantity with another. Example: Lets suppose y = 3x2. If you are asked to differentiate y with respect to x, then you are being asked to do: dy/dx. As you will soon learn, for the example above, dy/dx = 6x. The result or 6x is called the derivative and tells how y varies with x.

BASIC RULES OF DIFEFERNTIATION Just like in any other branch of Mathematics, there are basic rules that should be adhered to when differentiating. The rules below are the ones you may need to solve P14 problems. 1. The Power Rule This rule is used to find the derivative of a variable raised to a power. Example: Given y = x 4 , determine dy/dx. Rule: Multiply the variable (i.e. x in the example above) by the power it is raised to and subtract one from the power to get the new power of the variable.

So for the example, dy/dx = (4) x (4 -1 ) = 4 x 3

Another way of saying this is that the derivative of y = x4 is 4x3

Learning Points This document introduces you to the basics of differentiation. dy/dx means differentiate y with respect to x. Power Rule: Multiply the variable by the power it is raised to and subtract one from the power to get the new power of the variable.

2Lets practice this concept.

N 2 Tn R E

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T

Question 1

4 x2-1

Question 2 Given x = 4t

dx/dt = ( )4 t -1

= 2t -

Question 3 Given p = 4x2 + x-6

dp/dx = (2)4x2-1 + (-6)x-6-1 = 8x - 6x-7

Sample Problems Find the following derivatives:

a) dy/dx of y = 3x2 + 5x

b) dx/dt of x = 3t + 2t -2

Given y = 4x2 dy/dx = (2 ) = 8x

ow you try the sample problems on the right.

. The Constant Rule

his rule is used to find the derivative of a constant (otherwise known as a real umber).

ule: The derivative of a constant is zero.

xample : Given x = 9, then dx/dt = 0. Another way to say this is that the derivative of x = 9 is 0

ets practice this easy concept.

ry the sample problems on the right.

Question 4 Given y = 1/x This is the same as y = x-1 Therefore dy/dx = (-1)x-1-1

= -x-2

Question 5 Given y = x This is the same as y = x1 Therefore dy/dx = (1)x1-1

= x0 = 1

c) dv/dt of v = 5t3

d) du/dt of u = 6t3 - 8t

e) db/dt of b = 5t2 2t4 + 4t Answers a) 6x + 5 b) 3 - 4t -3 c) 15t2 d) 18t2 4t - e) 10t 8t3 + 4 Constant Rule: The derivative of a constant is zero. Sample Problems Find the following derivatives:

a) dy/dx of y = 3 + 5x-2

b) dx/dt of x = 3t + 2 Answers a) -10x-3 b) 3

Question 1 Given y = 143 dy/dx = 0

Question 2 Given p = 6 + 2x + 3x2 + 2x-6

dp/dx = 2 + 6x 12x-7

3

3. Differentiating Trigonometric Functions

Variables and real numbers are not the only things that we differentiate in this course. Trigonometric functions such as sine, cosine and tan are sometimes differentiated. Fortunately, you do not need to know how to differentiate these trigonometric functions. You only have to commit their derivates to memory.

Trigonometric Functions

dy/dx

y = sin x cos x y = cos x - sin x y = tan x sec2x

4. Exponential Functions When an exponential function is differentiated the exponential function is unchanged BUT it is multiplied by the differential of the exponent.

Example : Given y = 24te

Then dy/dt = 24te multiplied by 8t

= 8t24te

Lets practice these last two concepts.

5. Expressions raised to a power

So far we have been dealing with equations where variables are raised to a power e.g. y = x4. But what if we had an equation where an expression is raised to a power such as y = (2x + 5)5 or y = 5(4x2 + 6x)2. How do we even start to differentiate this type of equation? Do the following:

! Step 1: Apply the Power Rule treating the expression as a variable ! Step 2: Differentiate what is inside the brackets ! Step 3: Multiply the results from Step 1 and Step 2

4t2 is the exponent

Question 1 Given y = 3sinx + 4 cosx dy/dx = 3cosx 4sinx

Question 2 Given y = 4e3t

dy/dt = 4e3t (3) = 12 e3t

Know these derivatives by heart! Differentiating an exponential function gives back the function multiplied by the differential of the exponent.

4Example: Find dy/dx for y = (2x + 5)5

Step 1: Apply the Power Rule treating the expression as a variable. So imagine (2x + 5) to be a variable such as p. (It can be any variable you want). We are then differentiating y = p5. Using the Power Rule we would get 5p4. Therefore, from Step 1 we get 5(2x+5)4.

Step 2: Differentiate what is inside the brackets. 2x + 5 is the expression inside the brackets. Differentiating this with respect to x gives a answer of 2. Step 3: Multiply the results from Step 1 and Step 2. Result from Step 1 = 5(2x + 5)4 Result from Step 2 = 2 Multiplying the results from both steps = 5(2x + 5)4 (2) = 10 (2x + 5)4

Answer : dy/dx = 10 (2x + 5)4

Lets practice this concept. **Note** Use similar principles to do: dy/dx of y = sin(4x3).

! Step 1: Treat the expression in brackets as a variable and differentiate the trig function.

! Step 2: Differentiate what is inside the brackets ! Step 3: Multiply the results from Step 1 and Step 2

So for above:

Step 1: Imagine (4x3) to be a variable p. We are differentiating y = sin(p). We get cos(p) or cos(4x3).

Step 2: Differentiate what is inside the brackets. Differentiating 4x3 gives us 12x2 Step 3: Multiply the results from Step 1 and Step 2. dy/dx = cos(4x3) times 12x2 = 12x2cos(4x3)

Question 1 Given y = (2x2 + 4x)3

dy/dx = 3(2x2 + 4x)2 times (4x + 4) = 3(4x + 4) (2x2 + 4x)2

Question 2 Given p = 4(q3 2q-2)5

dp/dq = (5)4(q3 2q-2)4 times (3q2 + 4q-3) = 20(3q2 + 4q-3)(q3 2q-2)4

Step 1: Apply the Power Rule treating the expression as a variable.

Step 2: Differentiate what is inside the brackets.

Step 3: Multiply the results from Step 1 and Step 2.

Use similar principles for differentiating trig functions whose arguments are expressions.

5 Try the sample problems on the right. 6. The Product Rule Suppose you are given y = (x-2)(x2-1). This is a product equation of two parts (x-2) and (x2-1). We can apply the Product Rule to differentiate it. A good way to remember the Product Rule for differentiation is the first part times the derivative of the second part plus the second part times the derivative of the first part.

To put it a little simpler: 1st part x derivative of 2nd part + 2nd part x derivative of 1st part

So from above, if y = (x - 2)(x2- 1). dy/dx = (x - 2) times the derivative of (x2- 1)

+ (x2-1) times the derivative of (x-2) - 1)(1) L T 7 H SWtw = (x - 2)(2x) + (x2 = 2x2 - 4x + x2 - 1 = 3x2 - 4x -1

ets practice this concept. ry the sample problems on the right.

. One last thing - Differentiating with respect to something

opefully, you are becoming familiar with the rules of differentiation.

o far we have been looking at differentiating equations with one variable in it. hat if we had an equation with 2 or more variables (eg. p = x4y2 or k = 7y2t4)

o differentiate? Which variable/variables would we apply the Power Rule to and hat do we do with the rest of the equation?

Question 1 Given y = (6x3 -3x-2)(4x-2) dy/dx = (6x3 -3x-2)(4) + (4x-2)(18x2 +6x-3) = 24x3 12x-2 + 72x3 + 24x-2 -36x2 12x-3

= 96x3 -36x2 + 12x-2 12x-3

Question 2 Given y = 4x2 sinx dy/dx = 4x2(cosx) + sinx(8x) = 4x2cosx + 8xsinx Sample Problems Find dy/dx for the following :

a) y = (3x2 + 5)2 b) y = 2(5x 2)3 c) y = 3(4x5 + x)2 d) y = cos(x5)

Answers a) 12x(3x2 + 5) b) 30(5x 2)2 c) 6(4x5 + x)

(20x4 + 1) d) 5x4 sin(x5) Sample Problems

Find the dy/dx of the following using the Product Rule

a) y = (4x+5)(3x2+3) b) y = (3x2 + 7)2x c) y = (2x25)(2x6) d) y = 4x(2x2 13) Answers

a) 36x2 +30x +12b) 18x2 + 14 c) 12x2 - 24x 10 d) 24x2 52

6This is where we look to see what we are differentiating with respect to. When you know which variable you are differentiating with respect to, you can focus on applying the Power Rule to that variable. But theres still the question of what to do with the rest of the equation. Fortunately, there is a simple answer. Multiply the derivative of the variable youre differentiating with respect to by the rest of the equation. Rest of the equation means everything except the variable you applied the Power Rule to. Lets try this.

Example: Consider k = x4y2t6. Determine dk/dx In this example, the variable we are differentiating with respect to is x (because we are asked to do dk/dx). Treat everything else in the equation as a multiplied constant. So dk/dx = (4x3) y2t6

Lets practice this concept. T ight.

Conclusion Now you know how to differentiate (at least oming physics lectures, you will learn when to use it foyou think about it youve come a long way frothis will hard work will be wasted if you dPRACTICE, PRACTICE, PRACTICE. differentiation fresh in your mind.

The derivative of x4 is 4x3

Question 1 Given s = 3x2y, determine ds/dx ds/dx = 3y (2x) = 6xy

Question 2 Given s = 3x2y, determine ds/dy ds/dy = 3x2 (1) = 3x2 the basics). In your upcry the sample problems on the rr physics and where it applies. If m where you started. But all of o not practice to differentiate. Its the only way to keep dy/dx means differentiate y with respect to x. Sample Problems Find dy/dx for the following : a) y = 3x2t4 b) y = t4x c) y = s2x4t3 d) y = 2xt6s3 e) y = 3s2x2t5 f) y = 4s2t2x2 Answers a) 6xt4 b) t4 c) 4s2x3t3 d) 2t6s3 e) 6s2xt5 f) 8s2t2x