One Sheet Calculus Review
Derivatives
The Chain Rule
dun
= n un−1
du
df
df du
=
dx
du dx
deu
= eu
du
d sin(u)
= cos(u)
du
d cos(u)
= − sin(u)
du
d ln(u)
1
=
du
u
u-Substitution (Examples)
R
An application of the chain rule. We wish to do e−αt dt.
Let u = −αt. Thus du = −α dt. We convert the integral
into a u-form by muliplying and dividing by du
dt = −α to
turn dt into du:
Z
e−αt dt =
1
−α
Z
1
e−αt (−α dt) = − e−αt + C
α
Indefinite Integrals
Similarly:
Z
for (n 6= −1)
Z
for (n = −1)
Z
Z
Z
u
−1
un du =
Z
du =
un+1
+C
n+1
Z
du
= ln |u| + C
u
Z
eu du = eu + C
2
cos(ωt) dt =
(3x + 2) dx =
1
3
1
sin(ωt) + C
ω
1
(3x + 2)3
+ C = (3x + 2)3 + C
3
9
cos(u) du = sin(u) + C
Z
sin(u) du = − cos(u) + C
Note that in all cases a constant of integration must be
added if the integral is not used to evaluate a definite
integral (one with explicit limits, see next).
Definite Integral Rule
Given:
dv
= ln|v − mg/b| + C
v − mg/b
Taylor Series
Works best for “small” x:
d2 f x2
df x
+
+ ...
f (u) = f (a + x) = f (a) +
dx a
dx2 a 2!
dF (x)
= f (x)
dx
or
Binomial Expansion
dF = f (x) dx
A special case of the Taylor Series for f (u) = un =
(1 + x)n , expanded for u = 1 + x around 1. Requires
|x| < 1 to unconditionally converge:
then
F (x)|ba = F (b) − F (a) =
Z
b
dF =
a
Z
b
f (x) dx
a
(1 + x)n = 1 +
Integration by Parts
It’s differentiation of a product, both ways:
n does not have to be an integer in this expression!
d(uv) = v du + u dv
Move one term to the other side, rearrange, and integrate
both sides to get:
Z
u dv =
Z
d(uv) −
Z
v du = uv −
Z
nx n(n − 1)x2 n(n − 1)(n − 2)x3
+
+
+ ...
1!
2!
3!
v du