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