Post on 08-Mar-2018
Common Mistakes in CalculusS.-S. ChowLast revised Nov 4 1999
Forgetting to change sign
2x− 2(x + h) = 2h, x− (2y − 4) = x− 2y − 4
Forgetting constant factor
2x− 2(x + h) = −h,
Incorrectly combining terms
2x− 3x2 + πx = 2πx− 3x2
Ambiguous expression: incorrect use of parenthesis
1 + y/(1− y) =1 + y
1− y,
1 + x
x2 − 1= 1 + x/x2 − 1 = 1/x
log(x− 1)/x :log(x− 1)
xor log
x− 1x
log x− 1/x :log(x− 1)
xor log
(x− 1)x
or (log x)− 1x
orlog(x− 1)
x
Distributing power into expression
(a + b)2 = a2 + b2,√
x2 + 1 = x + 1
Transcription error
4x2 − 4x− 2 transcribed as 4x2 − 4x + 2, 4x2 + 4x + 2, or 4x2 − x + 2
(x2 − 4)x2 + 4x2 + 4
transcribed asx4 − 16x2 − 4
3x + sin(2x) transcribed as 3x + sin(x)
Deliberately or carelessly dropping term(s)
(4x2 − 4x− 2)(3x2 − 1) = (4x2 − 2)(3x2 − 1)
Factoring without cross checking:
x2 − 5x− 6 = (x− 3)(x− 2)
Inverting fraction2x + 3
(2x− 13 )
=13
2x + 3(6x− 1)
1
Partial cancellation of terms in fraction
a + b\b\
= a,a + b\c
b\= a + c
a\a\+ b
=1b,
a\c + b/d
a\+ b/= c + d
Break up of fractionc
a + b=
c
a+
c
b
Pulling out constant from trigonometric functions
sin(3x) = 3 sinx, sin(3x− π
4) = sin(3x)− sin(
π
4)
sin(a + b) = sin(a) + sin(b), sin(a + b) = sin a + cos b
sin(x− π
3) = sin x− sin
π
3Invalid values
cos3π
2= −1, tan
π
2= ∞
Incorrect trigonometric formulae
1sin(4x)
= cos(4x)
sin(3x) = 3 sinx cos x, cos(2x) = 2 cos x sinx
Confusing terms
(1− cos(4x))(1 + cos(4x)) = 1− cos(16x2)
Introducing new terms
x2 − 2x
sin(2x)=
x
sin(2x)· x
sin(2x)− 2x
sin(2x)
Misunderstanding notation
tan(3x) =sincos
(3x)
cos−1 x =1
cos xor cos
(1x
)tan−1 x =
sin−1 x
cos−1 x
Invalid valuese0 = 0, log 0 = 0
2
Assuming linear properties
e5x = 5ex, ex−2 = ex − e2
log 5x = 5 log x, log(a− b) = log a− log b
log2x
=log 2log x
Mistaking basic properties
log(a + b) = log a log b
log(a− b) =log a
log b
Left out limit symbol
limx→2
2x2 − 4x
x− 2= x
2x− 4x− 2
= 2x = 2
Correct answer but incorrect step(s)
1664
=1 6\6\ 4
=14
limx→0
x
sin 3x= lim
x→0
x
3 sinx=
13
limx→0
x
sinx=
13
Left out differentiation symbol
d
dx
1(x + 1/2)1/3
= (x + 1/2)−1/3 = −13(x + 1/2)−4/3
Keeping constant in differentiation
d
dx(sec x + 1) = sec x tanx + 1
d
dx(4x2 + 3)2 = 2(4x2 + 3)(8x + 3)
Apply wrong power ruled
dx2x = x2x−1
Mixing up power rule for integration and differentiation
d
dxx4 =
x5
5,
∫log x dx =
1x
Multiplying derivatives in product
d
dx(x3 + 1) sin(x) = 3x2 cos x
3
log y = log x log x2 + 2, so1y
dy
dx=
1x
1x2 + 2
(2x)
Applying wrong sign in quotient rule
d
dx
2x2 + 13x− 2
=(2x2 + 1)(3)− (3x− 2)(4x)
(3x− 2)2
Differentiating denominator and numerator separately in quotient (often due toconfusion with L’Hopital’s rule)
d
dx
x2 + 1sin(x)
=2x
cos x
d
dxtanx =
d
dx
sinx
cos x=
− cos x
sinx
Applying composite rule to all terms at once
d
dx(x + x2)3 = 3(1 + 2x)2,
d
dx
√x + x2 =
12(1 + 2x)−1/2
d
dxcos(3x) = − sin(3),
d
dxcos(3x2 + 4x) = sin(6x + 4)
Missing composite rule
d
dxe3x+4 = e3x+4,
d
dxlog(5x) = 5
1x
Mixing up integration and differentiation∫sinx dx = cos x + C,
∫cos x dx = − sinx + C,
∫tanx dx = sec2 x + C
Assuming integral∫
dx = 0:∫ 2
1
x cos x2+1 dx =∫ 2
1
x cos x2 dx+∫ 2
1
dx =12
∫ 2
1
(cos x2)(2x) dx =12
sin(x2)∣∣∣∣21
Left out integration symbol∫x(x + 2) dx = x2 + 2x =
x3
3+ x2 + C
Mixing up integration and differentiation∫x4 dx = 4x3 + C,
∫tanx dx = sec2 x + C
Mixing up definite integral and indefinite integral
d
dx
∫ 1
0
x3 − 3x dx = x3 − 3x∣∣x=1
x=0= −2
4
∫ 2
1
sec2 x dx = tan x|21 + C
Integrating product as product of integrals∫(x3 + 1) cos(x) dx =
∫x3 + 3 dx
∫sinx dx =
14x4 cos x + C
Integrating quotient as quotient of integrals∫x3 + 1cos(x)
dx =∫
x3 + 3 dx∫sinx dx
=x4/4cos x
+ C
Applying derivative to irrelevant term when using L’Hopital rule:
limx→0
log(1− x)x
+ 1 = limx→0
− 11−x
1+ 0
5