Common Algebraic and Calculus Errors

Introduction

The topics covered here are errors that students often make in doing algebra, and not just errors typically made in an algebra class. Typically, these mistakes are made by students in all level of classes, from algebra classes up to senior level math classes. Some of the mistakes in this article actually come from calculus courses.

Many of these are caused by student’s getting lazy or getting in a hurry and not paying attention to what they’re doing. By slowing down and paying attention to what you’re doing and paying attention to proper notation you can avoid the vast majority of these mistakes.

Many times students may understand the calculus concepts, but the algebra gets me every time. Having an efficient command of algebra is critical in mastering not only an algebra course but also in the study of calculus.

The article begins with common algebraic errors including trigonometry, logarithmic and functions. The article concludes with some common calculus errors including derivatives and integrals.

We begin with some common algebraic errors.

Algebraic Errors

\sf \frac{a}{x+b} \neq \frac{a}{x} + \frac{a}{b}

To see this error, a = b = x = 1.


\sf a - b(x - 1) \neq a - bx - b

Remember to distribute negative signs. The equation should be:

\sf a - b(x - 1) = a - bx + b


To divide fractions, invert and multiply.

\sf \frac{\left(\frac{x}{a}\right)}{b} \neq \frac{bx}{a}

The equation should be:

\sf \frac{\left(\frac{x}{a}\right)}{b} = \frac{\left(\frac{x}{a}\right)}{\left(\frac{b}{1}\right)} = \left(\frac{x}{a} \right) \left(\frac{1}{b} \right) = \frac{x}{ab}


\sf \frac{a + bx}{a} \neq 1 + bx

This is one of many examples of incorrect cancellations. The equation should be:

\sf \frac{a + bx}{a} = \frac{a}{a} + \frac{b}{x} = 1 + \frac{bx}{a}

\sf \frac{1}{x^{1/3} - x^{1/3}} \neq x^{-1/2} - x^{-1/3}

This error is a sophisticated version of the first error.


\sf (x^2)^3 \neq x^5

The equation should be:

\sf (x^2)^3 = x^2 x^2 x^2 = x^6


\sf (a + b)^2 \neq a^2 + b^2

The equation should be:

\sf (a + b)^2 = a^2 + 2ab + b^2


\sf |-2(x + 3)| \neq -2|x + 3|

The equation should be:

\sf |-2(x + 3)| = 2|x + 3|


\sf \sqrt{x + y} \neq \sqrt{x} + \sqrt{y}

There is no simplified way to write this expression.


\sf \sqrt{x^2+a^2} \neq x + a

To see this error, let x = 3 and a = 4


\sf \sqrt{-x^2 + a^2} = -\sqrt{x^2 - a^2}

We can’t factor a negative sign outside of the square root.


Solve for x:

\sf x(x - 2) = 1

\sf x = 1 or \sf x - 2 = 1

This is incorrect because the right hand side of the equation is not zero and the factors x and (x – 2) have been set equal to that non-zero number, that is, 1.

The zero product rule states that if ab = 0 then either a = 0 or b = 0. This rule is applicable only if the right hand side of the equation is zero.

\sf x(x - 2) = x^2 - 2x = 1

\sf x^2 - 2x - 1 = 0

Then try to factor, but since this will not factor nicely, use the quadratic formula to solve for x.

Trigonometric Errors

\sf \cos(10) = -0.839071529076 \text { radians}

\sf \cos(10) = 0.984807753013 \text { degrees}

Always use radians when dealing with trig functions.


\sf \cos(x + y) \neq \cos(x) + \cos(y)

For example, let \sf x = \pi and \sf y = 2\pi

\sf \cos(\pi + 2\pi) \neq \cos(\pi) + \cos( \pi)

\sf \cos(3\pi) \neq -1 + 1

\sf -1 \neq 0


\sf \cos(3x) \neq 3\cos(x)

For example, let \sf x = \pi

\sf \cos(3\pi) \neq 3\cos(\pi)

\sf -1 \neq 3(-1)

\sf -1 \neq -3

Remember: \sf \cos(x) is not multiplication; it represents the cosine function.


\sf \frac{\sin2x}{x} \neq\sin2

You can only cancel terms in the numerator and denominator of a fraction if they are not inside anything else and are just multiplying the rest of the numerator and denominator. The function \sf \sin2x is not \sf \sin2 multiplied by x. If the fraction had been written as:

\sf \frac{\sin(2x)}{x}

it would be harder to make such an error.


If \sf n is a positive integer then:

\sf \sin^2(x) = (\sin x)^n

However, the following two trig functions have different meanings:

\sf \tan^2x and \tan x^2

In the first case we are taking the tangent then squaring result and in the second we are squaring the \sf x then taking the tangent.


\sf \cos^{-1}(x) \neq \frac{1}{\cos(x)}

The -1 in the trig function \sf \cos^{-1}(x) is not an exponent, but to denote an inverse trig function.

Another notation for inverse trig functions that avoids this problem:

\sf \cos^{-1}(x) = \arccos(x)

However, this notation is seldom used.

Logarithmic Errors

Logarithmic term is expressed as \sf \log_aY , where the symbol a is known as the “base”.

If the base is 10, normally we will leave the logarithmic term as \sf \log Y , without writing the base 10. The examples shown use base 10 for simplicity.

The study of logarithms involves 3 basic laws. With these laws, any logarithmic expressions can be easily simplified. Here are the 3 laws:

1) Product Law:

\sf \log (XY) = \log X + \log Y the log terms are added.

2) Quotient Law:

\sf \log(\frac{X}{Y}) = \log X - \log Y the log terms are subtracted.

3) Power Law:

\sf \log X^n = n \log X the power \sf n is brought in front of the term.

Common mistakes:

\sf \log X + \log Y \neq \log (X + Y)

They are not equal.


Writing \sf \log (X/Y) as \log X / \log Y
It is “X divided by Y” before being “log”.

Thinking that \sf \log and X are separated; they are together as in: \sf \log X

Functional Errors

The exponent –1 should be treated carefully. When applied to a number, or an algebraic expression, it means the reciprocal. When applied to the notation of a function, it means the inverse.

For example, \sf 3^{-1} = \frac{1}{3} and \sf (x^2 + 2x + 1)^{-1}=\frac{1}{x^2 + 2x + 1}. But \sf \sin^{-1}(x) \neq \frac{1}{\sin(x)} and in general \sf f^{-1}(x) \neq \frac{1}{f(x)} therefore \sf \sin^{-1}x stands for the inverse sine function and \sf f^{-1}(x) stands for the inverse function of \sf f(x).

When a function \sf f(x) is raised to an exponent \sf k, \sf k \neq -1, it is written as \sf f^k(x). In other words, \sf f^k(x) = [f(x)]^k. This notation is especially common in trigonometry, e.g., \sf \sin^2x = [\sin x]^2 , \sf \tan^{1/2}x = \sqrt{\tan x}, etc.

Calculus Errors

\sf \frac{d}{dx} [2^x] \neq x2^{x-1}

The correct answer is \sf 2x\ln2 . The power rule only applies if the base is a variable and the exponent is a constant, as in \sf x^3 .


\sf \frac{d}{dx} [\sin(x^2+1)] \neq \cos(2x)

This is a typical example of the kind of mistakes made when applying the chain rule. The correct answer is:

\sf \frac{d}{dx} [\sin(x^2+1)] = \cos(x^2 + 1)\times 2x


\sf \frac{d}{dx} [\sin(x^2+1)] \neq \cos(x^2 + 1) + \sin(2x)

Another common way in which the chain rule is misapplied. This time the product rule has been used where the chain rule was the way to go.


\sf \frac{d}{dx} [\cos x] \neq \sin x

The answer should be \sf - \sin x . This is an extremely common error.


\sf \frac{d}{dx} \left[\frac{f}{g} \right]\neq \frac{fg^\prime - gf^\prime}{g^2}

This is backwards. It should be:

\sf \frac{d}{dx} \left[\frac{f}{g} \right]= \frac{gf^\prime - fg^\prime}{g^2}


\sf \frac{d}{dx}[\ln 3]\neq \frac13

The quantity \ln 3 is a constant, so \sf \frac{d}{dx}[\ln 3]= 0 . The same is true for all constants. So \sf \frac{d}{dx}[e]=0 and \sf \frac{d}{dx}[\sin \frac\pi2]= 0 as well.


\sf \int{x dx}\neq \frac{x^2}{2}

The correct answer is:

\sf \int{x dx} = \frac{x^2}{2} + C

Picky profs penalize points pedantically.


\sf \int{\frac{1}{x} dx} \neq \frac{x^0}{0} + C

The power rule for integration does not apply to \sf x^{-1}. Instead,

\sf \int{\frac{1}{x} dx} = \ln|x| + C


\sf \int{\tan x dx} \neq \sec^2 x + C

It is the other way around. \sf \frac{d}{dx}[\tan x] = \sec^2 x . The correct answer is:

\sf \int{\tan x dx} = \ln|\sec x| + C

as can be found by u-substitution with u = \cos x.


Forgetting to simplify mistake, for example:

\sf \int{x\sqrt x dx}

is easy if you notice that \sf x\sqrt x = x^\frac23 and then apply the power rule for integration. If you try to do this problem by using integration by parts or substitution, you will find this approach much more difficult, if not impossible to solve.


Not substituting back to the original variable mistake

\sf \int{2xe^{x^2} dx}

does not equal \sf e^u + C . It equals \sf e^{x^2} + C .

Conclusion

The worst mistake many students make is to think they know the material better than they really do. It is easy to fool yourself into thinking you can solve a problem when you’re looking at the answer book or at a worked out solution. Test your knowledge by trying problems under exam conditions. If you can do them under that restriction, the exam should be a breeze.

Do not fear mistakes. We can learn from these mistakes, but, after looking and understanding these mistakes, we should correct and not make them again.

We hope you found this article informative and wish the best to your child’s success.

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