17Calculus Precalculus - Order of Operations
In algebra, you probably ran across the concept of 'Order Of Operations'. It will help you a lot if you take a minute and refresh your memory on this concept. It is critical to your success in calculus mostly because some calculus classes use calculators and computers which require you enter your expressions using these rules.
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Here is the correct order of operations from highest to lowest priority.
1. Parentheses |
2. Exponents and Radicals |
3. Multiplication and Division (left to right) |
4. Addition and Subtraction (left to right) |
This list means that the strongest binding is parentheses. Parentheses override all other operations. Secondly, exponents and radicals overide operations 3 and 4, etc.
IMPORTANT: You need to work left to right within rules 3 and 4. Working right to left may give you the wrong answer.
Here are some examples.
Example 1
\(\displaystyle{ x + 3 / 2 = x + \frac{3}{2} }\)
If you want the entire x + 3 expression to be divided by 2, you need to use parentheses to override rule 3 taking precedence to rule 4.
\(\displaystyle{ (x + 3) / 2 = \frac{x+3}{2} }\)
Example 2
e^2x = \( e^2x \)
If you want 2x to be the exponent to e, you need to use parentheses to override rule 2 taking precedence to rule 3. e^(2x) = \( e^{2x} \)
Example 3
Here is one I see a lot. A student enters -3^2 into their calculator. This evaluates to -9 because the 2 power takes precedence to the negative sign. So this is like entering -(3^2). The negative sign falls in the same category as subtraction. We just usually do not write 0-3^2. To get (-3)^2, the parentheses are required.
Note: Take special care not to use computer notation in your written work. Good teachers require correct notation and computer notation is not considered correct notation for written work.
Acronym PEMDAS
It is okay to use the acronym PEMDAS as long you remember that multiplication (M) and division (D) are actually one rule. Similarly for addition and subtraction. So, you should think about this acronym as PE(MD)(AS) to give you four rules. Just remember to always to left-to-right within the rules.
Operation Stickiness
In order to avoid making mistakes within rules 3 and 4, there is another way to write the equations. I will show you how I think of it using an example.
Example 4
\( 3 \div 4 \times 16 \)
I think of the operation before a number (or variable) as being stuck to the number. Then I convert the division into multiplication. So to divide by 4 is the same multiplying by 1/4. The equation is now \(\displaystyle{ 3 \times \frac{1}{4} \times 16 }\). Now I just multiply everywhere and I can do it in any order. It makes more sense to me to multiply the last two since I will get an integer. So I have \( 3 \times 4 = 12 \). I can do this only because I replaced the division by multiplication. This technique will come in handy in calculus.
Example 5
Now let's look at addition and subtraction. Again, I think of the operation as being stuck to the number it follows. So in \( 3 - 4 + 16 \) the subtraction sign in front of the 4 is stuck to the 4. I can move the subtraction and replace it with addition if I stick a negative sign to the 4.
\( 3 + (-4) + 16\) (The parentheses are required. 3 + - 4 does not make sense.) Now I have addition everywhere and I can do the additions in any order.
\( -1+16 = 3+12 = 15 \)
Here is a video that is worth watching explaining how PEMDAS is not the best way to understand the math behind the order of operations.
minutephysics - The Order of Operations is Wrong
video by minutephysics |
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Okay, so try the practice problems and make sure you are confident with evaluating expressions. Calculus absolutely requires you to know these well. If you have trouble with these problems, here are a few websites for you to go and review.
This page has a great discussion of PEMDAS and how to avoid errors. Math Is Fun: Order of Operations: PEMDAS |
Practice
Unless otherwise instructed, simplify these expressions using the correct order of operations.
Simplify \( 12 - 2 \times 4 + 9 \)
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Simplify \( 12 - 2 \times 4 + 9 \)
Final Answer |
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\( 13 \)
Problem Statement |
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Simplify \( 12 - 2 \times 4 + 9 \)
Solution |
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2701 video
Final Answer |
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\( 13 \) |
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Simplify \( 6 \div 2 + 3[ 15 + 3(7-9)^3 ] \)
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Simplify \( 6 \div 2 + 3[ 15 + 3(7-9)^3 ] \)
Final Answer |
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\( -24 \)
Problem Statement |
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Simplify \( 6 \div 2 + 3[ 15 + 3(7-9)^3 ] \)
Solution |
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2702 video
Final Answer |
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\( -24 \) |
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Simplify \( 20 \div 5 + 4^2 [ (-3+7)-1]-2^3 \)
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Simplify \( 20 \div 5 + 4^2 [ (-3+7)-1]-2^3 \)
Final Answer |
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\( 44 \)
Problem Statement |
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Simplify \( 20 \div 5 + 4^2 [ (-3+7)-1]-2^3 \)
Solution |
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2703 video
Final Answer |
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\( 44 \) |
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Simplify \( 7\sqrt{9} + 3 - \sqrt{64} \)
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Simplify \( 7\sqrt{9} + 3 - \sqrt{64} \)
Final Answer |
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\( 15 \)
Problem Statement |
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Simplify \( 7\sqrt{9} + 3 - \sqrt{64} \)
Solution |
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2704 video
Final Answer |
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\( 15 \) |
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Simplify \( 8 - 2[ 3 - 4\sqrt{50-14} + 5] \)
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Simplify \( 8 - 2[ 3 - 4\sqrt{50-14} + 5] \)
Final Answer |
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\( 40 \)
Problem Statement |
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Simplify \( 8 - 2[ 3 - 4\sqrt{50-14} + 5] \)
Solution |
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2705 video
Final Answer |
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\( 40 \) |
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Simplify \(\displaystyle{ \frac{4(6+4)+(7-5)^5}{6(4-2)-2^2} }\)
Problem Statement |
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Simplify \(\displaystyle{ \frac{4(6+4)+(7-5)^5}{6(4-2)-2^2} }\)
Final Answer |
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\( 9 \)
Problem Statement |
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Simplify \(\displaystyle{ \frac{4(6+4)+(7-5)^5}{6(4-2)-2^2} }\)
Solution |
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2706 video
Final Answer |
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\( 9 \) |
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Simplify \(\displaystyle{ \frac{3\cdot 4^2 - 3^2 \cdot 5}{5\cdot 9} }\)
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Simplify \(\displaystyle{ \frac{3\cdot 4^2 - 3^2 \cdot 5}{5\cdot 9} }\)
Final Answer |
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\( 1/15 \)
Problem Statement |
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Simplify \(\displaystyle{ \frac{3\cdot 4^2 - 3^2 \cdot 5}{5\cdot 9} }\)
Solution |
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2707 video
Final Answer |
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\( 1/15 \) |
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Simplify \( 42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) } \)
Problem Statement |
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Simplify \( 42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) } \)
Final Answer |
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\( 147 \)
Problem Statement |
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Simplify \( 42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) } \)
Solution |
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2708 video
Final Answer |
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\( 147 \) |
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Simplify \(\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }\)
Problem Statement |
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Simplify \(\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }\)
Final Answer |
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\( 43.5 \)
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Simplify \(\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }\)
Solution |
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2709 video
Final Answer |
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\( 43.5 \) |
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Really UNDERSTAND Precalculus
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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