B Veitch
Calculus 2 Derivative and Integral Rules
1. Derivative Formulas
(a) Common Derivatives
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
xii.
xiii.
xiv.
d
(c) = 0
dx
d
(f ± g) = f 0 ± g 0
dx
d
(x) = 1
dx
d
(kx) = k
dx
d
Power Rule:
(xn ) = nxn−1
dx
d
([f (x)]n ) = n[f (x)]n−1 · f 0 (x)
dx
Product Rule: (f g)0 = f 0 g + f g 0
0
f
f 0 g − f g0
Quotient Rule:
=
g
g2
0
0
Chain Rule: (f (g(x))) = f (g(x)) · g 0 (x)
d x
(a ) = ax ln a
dx
d x
(e ) = ex
dx
d
1
(ln x) =
dx
x
d
1
(loga x) =
dx
x ln a
d f (x) e
= f 0 (x) · ef (x)
dx
xv.
xvi.
xvii.
xviii.
xix.
xx.
xxi.
xxii.
xxiii.
xxiv.
xxv.
xxvi.
xxvii.
d
1
(ln(f (x))) =
· f 0 (x)
dx
f (x)
d
(sin x) = cos x
dx
d
(cos x) = − sin x
dx
d
(tan x) = sec2 x
dx
d
(sec x) = sec x tan x
dx
d
(csc x) = − csc x cot x
dx
d
(cot x) = − csc2 x
dx
1
d
sin−1 x = √
dx
1 − x2
d
−1
cos−1 x = √
dx
1 − x2
d
1
tan−1 x =
dx
1 + x2
−1
d
cot−1 x =
dx
1 + x2
1
d
sec−1 x = √
dx
x x2 − 1
d
−1
csc−1 x = √
dx
x x2 − 1
2. L’Hospitals Rule
(a) If lim f (x) = 0 and lim g(x) = 0, then
x→a
x→a
f 0 (x)
f (x)
= lim 0
, g 0 (x) 6= 0
x→a g (x)
x→a g(x)
lim
(b) If lim f (x) = ±∞ and lim g(x) = ±∞, then
x→a
x→a
f (x)
f 0 (x)
= lim 0
x→a g(x)
x→a g (x)
lim
(c) Indeterminate Form of Type 0 · ∞
Given lim f (x)g(x), you do the following to put it in the form
x→a
f (x)
1/g(x)
g(x)
ii. Rewrite f (x)g(x) as
1/f (x)
i. Rewrite f (x)g(x) as
(d) Indeterminate Form of 00 , ∞0 , or 1∞
Given lim f (x)g(x) , rewrite the limit as
x→a
lim eg(x) ln(f (x))
x→a
1
0
∞
or
0
∞
B Veitch
Calculus 2 Derivative and Integral Rules
then take the limit of the exponent
lim g(x) ln(f (x))
x→a
This should put the limit in the Indeterminate Form of Type 0 · ∞
3. Integrals
(a) Common Integrals
Z
i.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
xii.
xiii.
xiv.
Z
xn+1
x dx =
+ C, n 6= −1
n+1
Z
1
dx = ln |x| + C
Z x
1
1
dx = ln |kx + b| + C
kx
+
b
k
Z
ex dx = ex + C
Z
1
ekx dx = ekx + C
k
Z
1
kx+b
e
dx = ekx+b + C
k
Z
ax
ax dx =
+C
ln a
Z
1
akx dx =
akx + C
k
ln
a
Z
1
akx+b dx =
akx+b + C
k
ln
a
Z
ln x dx = x ln x − x + C
Z
cos x dx = sin x + C
Z
sin x dx = − cos x + C
Z
ii.
Z
k dx = kx + C
n
xv.
sec2 x dx = tan x + C
csc2 x dx = − cot x + C
Z
xvi.
sec x tan x dx = sec x + C
Z
csc x cot x dx = − csc x + C
xvii.
Z
tan x dx = ln | sec x| + C
xviii.
Z
cot x dx = ln | sin x| + C
xix.
Z
csc x dx = ln | csc x − cot x| + C
xx.
Z
xxi.
Z
xxii.
Z
xxiii.
Z
xxiv.
Z
xxv.
Z
xxvi.
sec x dx = ln | sec x + tan x| + C
x
1
√
dx = sin−1
+C
a
a2 − x2
1
1
−1 x
dx
=
tan
+C
a2 + x2
a
a
1
1
x
− 2
dx = cot−1
+C
2
a +x
a
a
1
√
dx = sec−1 (x) + C
x x2 − 1
1
dx = csc−1 (x) + C
− √
x x2 − 1
4. Integration Techniques
(a) u Substitution
Z b
Given
f (g(x))g 0 (x) dx,
a
i. Let u = g(x)
ii. Then du = g 0 (x) dx
iii. If there are bounds, you must change them using u = g(b) and u = g(a)
Z b
Z g(b)
f (g(x))g 0 (x) dx =
f (u) du
a
g(a)
(b) Integration By Parts
Z
Z
u dv = uv −
Z
Example:
x2 e−x dx
2
v du
B Veitch
Calculus 2 Derivative and Integral Rules
dv = e−x dx
u = x2
du = 2x dx
v = −e−x
Z
Z
x2 e−x dx = −x2 e−x − −2xe−x dx
You may have to do integration by parts more than once. When trying to figure out what to
choose for u, you can follow this guide: LIATE
L Logs
I Inverse Trig Functions
A Algebraic (radicals, rational functions, polynomials)
T Trig Functions (sin x, cos x)
E Exponential Functions
(c) Products of Trig Functions
Z
i.
Z
sinn x cosm x dx
ii.
A. m is odd (power of cos x is odd). Factor out one cos x and place it in front
of dx. Rewrite all remaining cos x as
sin x by using cos2 x = 1−sin2 x. Then
let u = sin x and du = cos x dx
tann x secm x dx
A. m is even (power of sec x is even). Factor out one sec2 x and place it in front
of dx. Rewrite all remaining sec x as
tan x by using sec2 x = 1 + tan2 x.
Then let u = tan x and du = sec2 x dx
B. n is odd (power of sin x is odd). Factor out one sin x and place it in front
of dx. Rewrite all remaining sin x as
cos x by using sin2 x = 1−cos2 x. Then
let u = cos x and du = − sin x dx
B. n is odd (power of tan x is odd). Factor out one sec x tan x and place it
in front of dx.
Rewrite all remain-
ing tan x as sec x by using tan2 x =
sec2 x − 1. Then let u = sec x and
C. If n and m are both odd, you can
du = sec x tan x dx
choose either of the previous methods.
D. If n and m are even, use the following
C. If n odd and m is even, you can use
either of the previous methods.
trig identities
1
(1 − cos(2x))
2
1
cos2 x = (1 + cos(2x))
2
sin(2x) = 2 cos x sin x
D. If n is even and m is odd, the previous
sin2 x =
methods will not work. You can try to
simplify or rewrite the integrals. You
may try other methods.
(d) Partial Fractions
Z
p(x)
dx, the degree of p(x) must be smaller than
q(x)
the degree of q(x). Factor the denominator q(x) into a product of linear and quadratic factors.
Use this method when you are integrating
There are four scenarios.
3
B Veitch
Calculus 2 Derivative and Integral Rules
Unique Linear Factors: If your denominator has unique linear factors
x
A
B
=
+
(2x − 1)(x − 3)
2x − 1 x − 3
To solve for A and B, multiply through by the common denominator to get
x = A(x − 3) + B(2x − 1)
1
. Find B by plugging in x = 3.
2
Repeated Linear Factors: Every power of the linear factor gets its own fraction (up to the
You can find A by plugging in x =
highest power).
x
A
D
B
C
=
+
+
+
(x − 3)(3x + 4)3
x − 3 3x + 4 (3x + 4)2
(3x + 4)3
Unique Quadratic Factor:
3x − 1
A
Bx + C
=
+ 2
2
(x + 4)(x + 9)
x+4
x +9
To solve or A, B, and C, multiply through by the common denominator to get
3x − 1 = A(x2 + 9) + (Bx + C)(x + 4)
Repeated Quadratic Factor: Every power of the quadratic factor gets its own fraction (up
to the highest power).
2x
A
Bx + C
Dx + E
Fx + G
=
+ 2
+ 2
+ 2
(3x + 4)(x2 + 9)3
3x + 4
x +9
(x + 9)2
(x + 9)3
(e) Trig Substitution
If you see
√
a2 − x2
√
a2 + x2
√
x2 − a2
Z
x3
√
Example:
dx Let
16 − x2
Z
√
x3
dx =
16 − x2
Substitute
x = a sin(θ)
1 − sin2 (θ) = cos2 (θ)
x = a tan(θ)
1 + tan2 (θ) = sec2 (θ)
x = a sec(θ)
sec2 (θ) − 1 = tan2 (θ)
x = 4 sin θ, dx = 4 cos θ dθ. So x3 = 64 sin3 θ
64 sin3 θ
Z
p
Uses the following Identity
Z
2
· 4 cos θ dθ =
16 − 16 sin θ
64 sin3 θ · 4 cos θdθ
=
4 cos θ
Z
sin3 θ dθ
Now you have an integral containing powers of trig functions. You can refer to that method to
solve the rest of this integral.
Z
sin3 dθ =
cos3 θ
− cos θ + C
3
To get back to x, we need to use a right triangle with the original substitution x = 4 sin x.
4