Differentiation and Integration Rules

Differentiation and Integration Rules
BCCC Tutoring Center
This handout highlights some of the most frequently encountered rules for
differentiation and integration.
For the following let u and v be functions of x, let n be an integer, and
let a, c and C be constants.
Fundamental Rules
d(c)
=0
dx
R
du
d(cu)
=c
dx
dx
R
d(u + v)
du dv
=
+
dx
dx dx
R
R
R
v
d( uv )
=
dx
du
dx
−u
v2
cdu = c
(du + dv) =
du
d(un )
= nun−1
dx
dx
du
dv
d(uv)
=v
+u
dx
dx
dx
du = u + C
R
un du =
R
du + C
du +
R
dv + C
un+1
+C
n+1
u dv = uv −
R
v du + C
dv
dx
Trigonmetric Functions
d(sin u)
= cos u
dx
R
du
dx
d(cos u)
du
= − sin u
dx
dx
sin u du = − cos u + C
R
cos u du = sin u + C
d(tan u)
du
= sec2 u
dx
dx
R
tan u du = ln | sec u| + C
d(cot u)
du
= − csc2 u
dx
dx
R
cot u du = ln | sin u| + C
d(sec u)
du
= sec u tan u
dx
dx
R
sec u du = ln | sec u + tan u| + C
Trigonmetric Functions Continued
d(csc u)
du
= − csc u cot u
dx
dx
R
R
R
csc u du = − ln | csc u + cot u | + C
R
sec2 u du = tan u + C
R
sec u tan u du = sec u + C
csc2 u du = − cot u + C
csc u cot u du = − csc u + C
Exponential and Logarithmic Functions
d(eu )
du
= eu
dx
dx
R
eu du = eu + C
R 1
du = ln |u| + C
u
d(ln u)
1 du
=
dx
u dx
d(au )
du
= (au )(ln a)
dx
dx
R
au du =
au
+C
ln a
v
d(uv )
= uv (v 0 ln u + u0 )
dx
u
Inverse Trigonometric Functions
d(sin−1 u)
1
du
=√
2
dx
1 − u dx
Z
d(cos−1 u)
−1
du
=√
dx
1 − u2 dx
Z
d(tan−1 u)
1
du
=
2
dx
1 + u dx
d(csc−1 u)
−1
du
= √
dx
|u| u2 − 1 dx
√
−1
u
= − sin−1 + C
2
2
a
a −u
Z
a2
d(cot−1 u)
−1 du
=
dx
1 + u2 dx
d(sec−1 u)
1
du
= √
2
dx
|u| u − 1 dx
1
u
√
= sin−1 + C
a
a2 − u 2
Z
Z
Z
1
1
u
= tan−1 + C
2
+u
a
a
−1
1
−1 u
=
−
tan
+C
a2 + u 2
a
a
1
1
u
√
= sec−1 | | + C
2
2
a
a
|u| u − a
1
u
−1
√
= − sec−1 | | + C
a
a
|u| u2 − a2
Hyperbolic Trigonometric Functions
d(sinh u)
du
= cosh u
dx
dx
R
sinh u du = cosh u + C
du
d(cosh u)
= sinh u
dx
dx
R
cosh u du = sinh u + C
d(tanh u)
du
= sech2 u
dx
dx
R
tanh u du = − ln |sech u | + C
Inverse Hyperbolic Trigonometric Functions
1
du
d(sinh−1 u)
=√
dx
1 + u2 dx
R
√
1
d(cosh−1 u)
du
=√
2
dx
u − 1 dx
R
√
1
du
d(tanh−1 u)
=
2
dx
1 − u dx
R
3
1
du = sinh−1 u + C
1 + u2
1
u2
−1
du = cosh−1 u + C
1
du = tanh−1 u + C
2
1−u