This section lists common mistakes that students can sometimes make. For ease of reference the errors are divided into four main categories. The Calculus Errors section deals with errors associated with derivatives and integrals. The Root and Exponent Errors section deals with errors involving exponents and roots. The Algebraic Manipulation Errors section includes errors which may occur when dealing with manipulating expressions. Finally, the Trigonometric Errors section deals with common errors involving trigonometry.

Note that these are listed by ERROR so the formula that appears above the explanation is INCORRECT. The correct formula (if one exists) appears at the end of the explanation.

Calculus Errors

[f(x)g(x)]' = [f '(x)][g '(x)]

Let us assume that this is true. Let f(x)=x and g(x)=2x. Then we have:

[f(x)g(x)]' = [f '(x)][g '(x)]
[x(2x)]' = [(x)'][(2x)']
[2x²]' = 1(2)
4x = 2

This example clearly shows the error in assuming that [f(x)g(x)]' = [f '(x)][g '(x)] since the incorrect answer (the right side) is not equal to the correct answer (the left side). The correct method to be used is the Product Rule where [f(x)g(x)]' = [f '(x)][g(x)] + [f(x)][g '(x)].

[f(x) / g(x)]' = [f '(x)] / [g '(x)]

Let us assume that this is true. Let f(x)=x and g(x)=2x. Then we have:

[f(x) / g(x)]' = [f '(x)] / [g '(x)]
[x / 2x]' = (x)' / (2x)'
[1 / 2]' = 1 / 2
0 = 1 / 2

From this example it can be seen that [f(x)g(x)]' = [f '(x)][g '(x)] is the incorrect. The correct method to be used is the Quotient Rule where [f(x) / g(x)]' = ( [f '(x)][g(x)] - [f(x)][g '(x)] ) / [g(x)]².

Let us assume that this is true. Let f(x)=x and g(x)=2x. Then we have:



From this it is easy to see that evaluating the integral of a product as the product of two integrals gives the wrong answer. The correct method to use here is the Substitution Rule (unless the integral can be simplified and directly evaluated as with the above example).

Note: The arbitrary constant C is the same on both sides since the assumption was that both sides of the equation were equal, hence the constant is assumed to be the same on both sides.

Let us assume that this is true. Let f(x)=x and g(x)=2x. Then we have:



As can be seen the answer on the right is not equal to the answer on the left (which is the correct answer) so this way of evaluating the integral of a quotient is incorrect. The correct method to be used is the Substitution Rule (unless the integral can be simplified and directly evaluated as with the above example).

Note: The arbitrary constant C is the same on both sides since the assumption was that both sides of the equation were equal, hence the constant is assumed to be the same on both sides.

If we were to use the Power Rule as this mistake seems to suggest, we would actually have:

Since division by 0 is undefined this method is obviously incorrect. The correct answer to the integral is ln|x|+C.

Root and Exponent Errors

Let us assume that this is true. Let a=1 and b=4. Then we have:



Since the square root of 5 is not equal to three we can see that you cannot split the root in such a way.

Let us assume that the equation is true. Let a=-1 and b=-4. Then we have:



This example made use of complex numbers. It can be shown that with the use of complex numbers either a or b can be negative but not both. However in MATH 1036/1037 we are only looking for real valued answers so for this course, both a and b must be positive.

Let us assume that the equation is true. Let a=-4. Then we have:



This example made use of complex numbers. Since in MATH 1036/1037 only real valued answers are desired and (as shown in the example) the negative sign cannot be removed from the square root, the correct response in this course would be "no real valued answers exist".

(cx)ⁿ = cxⁿ where c is constant.

Let us assume that the equation is true. Let c=2, x=3, and n=2. Then we have:

(cx)ⁿ = cxⁿ
((2)(3))² = 2(3²)
6² = 2(9)
36=18

It is quite clear that 36 is not equal to 18 so the assumption that this is true is wrong. The correct formula for removing brackets in this case is (cx)ⁿ = cⁿxⁿ.

Let us assume that the equation is true. Let c=8, x=8, and n=3. Then we have:



As 4 is not equal to 16 this formula is flawed.  The correct formula is .

(x+y)² = x²+y²

Let us assume that the equation is true. Let x=2 and y=4. Then we have:

(x+y)² = x²+y²
(2+4)² = 2²+4²
6² = 4+16
36 = 20

Since 20 is not equal to 36 (36 being the correct answer) this formula is wrong. The correct formula is (x+y)² = x²+2xy+y² because (x+y)² means (x+y)(x+y).

(-x)² = -x²

Let us assume that the equation is true. Let x=2. Then we have:

(-x)² = -x²
(-2)² = -2²
4 = -(2²)
4 = -4

This example shows that the negative sign cannot be factored from the brackets when raised to the power 2. The correct way to solve this is (-x)² = x².  Note that this property holds for any even power but if the exponent is odd the negative sign can be factored as in the example.

Let us assume that the equation is true. Let x = -1. Then we have:



The correct way to simplify this is:

Algebraic Manipulation Errors

x-(a-b) = x-a-b

Let us assume that this is true. Let x=2, a=3, and b=4. Then we have:

x-(a-b) = x-a-b
2-(3-4) = 2-3-4
2-(-1)= -5
2+1 = -5
3 = -5

As can be seen -5 is not equal to 3 so the formula is incorrect.  The error was that the negative sign must be multiplied into both terms in the bracket leaving x-(a-b) = x-a+b.

(ax²+b) / x = ax+b

Let us assume that this is true. Let x=2, a=3, and b=4. Then we have:

(ax²+b) / x = ax+b
(3(2²)+4) / 2 = 3(2)+4
(3(4)+4) / 2 = 6+4
16 / 2 = 10
8 = 10

Since 10 does not equal 8 this formula is incorrect.  The error was believing that in dividing by x you only needed to remove x from one term in the numerator. In order to do this EVERY term in the numerator must have an x in it.

ax² = x solved by dividing both sides by x.  I.E. solve ax = 1

Let us assume that this is true. Let  a=2. Then we have:

ax² = x
ax = 1
2x = 1
x = 1/2

This seems plausible since we get a value for x and when substitute into the equation it works. However, if we solve it the right way (by subtracting x from both sides so the equation is equal to zero) we will see why dividing by x is not the way to solve the equation.

ax² = x
ax²-x = 0
x(ax-1) = 0
x(2x-1) = 0

so x = 0 or 2x-1 = 0

x = 0 or 2x-1 = 0
x = 0 or 2x = 1
x = 0 or x = 1/2

Therefore this method gives x = 1/2 as an answer as well but it also gives x = 0 as another possible value for x. This means that the first method is incorrect and the correct way to solve the equation is to subtract x from both sides.

1 / (x+y) = 1/x + 1/y

Let us assume that this is true. Let x=2 and y=3. Then we have:

1 / (x+y) = 1/x + 1/y
1 / (2+3) = 1/2 + 1/3
1/5 = 3/6 + 2/6
1/5 = 5/6

Which is obviously false. It is often better to leave terms of this type in their original form as splitting them up is usually more difficult than evaluating directly.

Trigonometric Errors

sin(x) = sin(x) (note that the symbol beside the second x is degree not power 0)

In calculus sin(x) means the sine of x radians NOT x degrees.  To show an example that these are different, let x=2.  Then we have:

sin(x) = sin(2) 0.03489950
sin(x) = sin(2) 0.90929743

Even though the value of 2 was used in both equations, the difference between 2 degrees and 2 radians made the answers different. Remember that in calculus the value will almost always be in RADIANS.

This applies to all trigonometric terms, not just sin(x).

sin(x+y) = sin(x) + sin(y)

Let us assume that this is true. Let x= and y=/6. Then we have:

sin(x+y) = sin(x) + sin(y)
sin(+/6) = sin() + sin(/6)
sin(7/6) = 0+1/2
-1/2 = 1/2

This shows that the assumed formula is incorrect (since 1/2 is not equal to -1/2).  The correct formula in this situation is

sin(x+y) = sin(x)cos(y)+cos(x)sin(y). Formula for the other trigonometric functions can be found in the Trigonometric Formulas section.

Note that this applies to any trigonometric function and also for '-' instead of '+'.

cos^-1(x) = 1 / cos(x)

Let us assume that this is true. Let x=0. Then we have:

cos^-1(x) = 1 / cos(x)
cos^-1(0) = 1 / cos(0)
/2 = 1 / 1
/2 = 1

Since /2 is not equal to 1 so the assumption is wrong. This is a common mistake since people often think that because cos²(x) = (cos(x))² then cos^-1(x) should equal (cos(x))^-1. However, cos^-1(x) denotes the inverse cosine function, also called arccosine, not the cosine function raised to the power -1. See the Inverse Trigonometric Functions section for more details.

Note that this observation applies for any trigonometric function.

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