Some Dif f erentiation Rules
The following pages list various rules for finding derivatives with very basic examples to show how the
rules are used.
Note: The following pages are NOT formula sheets for exams or quizzes.
Definition: Let f (x) be a function. Then the derivative of f (x) is the function denoted f 0 (x) given by
f 0 (x) = lim
h→0
f (x + h) − f (x)
h
PROVIDED this limit exists.
For a particular value of x, say x = a, the derivative evaluated at x = a is given by
f 0 (a) = f 0 (x)
x=a
f (a + h) − f (a)
h→0
h
= lim
PROVIDED this limit exists at x = a.
A function y = f (x) is differentiable at a if f 0 (a) exists, i.e., if the above limit exists. This value f 0 (a)
is called the derivative of f at x = a.
Other notations: f 0 (x)
df
dx
d
f
dx
Dx f
y0
Dx(2) f
y 00
Dx(3) f
y 000
Dx(4) f
y (4)
Dx(n) f
y (n)
dy
dx
d
y
dx
Dx y
Notations for Higher Order Derivatives:
2nd order:
f 00 (x)
3rd order:
f 000 (x)
4th order:
f (4) (x)
nth order:
f (n) (x)
d2 f
dx2
d3 f
dx3
d4 f
dx4
dn f
dxn
d2
f
dx2
d3
f
dx3
d4
f
dx4
dn
f
dxn
d2 y
dx2
d3 y
dx3
d4 y
dx4
dn y
dxn
d2
y
dx2
d3
y
dx3
d4
y
dx4
dn
y
dxn
Dx(2) y
Dx(3) y
Dx(4) y
Dx(n) y
The point slope form of the equation of a line with slope m and through the point (x1 , y1 ) is given
by y − y1 = m(x − x1 ). We can use this form of the equation of a line to express the equation of the
Tangent Line to a curve y = f (x) at the point (a, f (a)) by the equation y − f (a) = f 0 (a)(x − a).
The slope of the tangent line to the curve a the point (a, f (a)) is given by the value of the derivative
f 0 (x) evaluated at a; so m = f 0 (a). At the point (a, f (a)), x1 = a, and y1 = f (a).
The normal line to a curve y = f (x) at the point (a, f (a)) is perpendicular to the tangent line to the
0
curve at that same point. It is given by y − f (a) = f−1
0 (a) (x − a) (provided f (x) 6= 0).
The derivative gives the instantaneous rate of change of y = f (x) with respect to x at the instant
x = a.
If y = f (x) gives the position of a moving particle, then the first derivative f 0 (a) gives the instantaneous velocity of the particle at the instant x = a while the second derivative f 00 (a) gives the
instantaneous acceleration of particle at the instant x = a.
Example: y = f (x) = x3 + 5x2 , f 0 (x) = 3x2 + 10x. When x = 2, y = f (2) = 28, then f 0 (2) = 32 and
the equation of the tangent line y − f (a) = f 0 (a)(x − a) becomes y − 28 = 32(x − 2).
1
Some Dif f erentiation Rules
Dif f erentiation Rules
b and c are real constants; b > 0, b 6= 1
Examples
√
( 5)0 = 0
(6)0 = 0
1. (c)0 = 0
0
0
(π 2 )0 = 0
0
2. (cx) = c
x =1
3. (xn )0 = nxn−1 n = any real number
(x8 )0 = 8x7
0
(6x) = 6
1
x5
(πx) = π
0
= x−5
√
1
1 2
( 3 x)0 = (x 3 )0 = x− 3
3
4. Sum & Difference Rule:
[f (x) ± (x)]0 = f 0 (x) ± g 0 (x))
5.
cf (x)
0
6. (ex )0 = ex
(5x2 + ex )0 = 10x + ex
7. (bx )0 = bx ln b
(6x)0 = 6x ln 6
8. (ln x)0 =
1
x
9. (logb x)0 =
(3 + 2x + ln x)0 = 2 +
1
x ln b
= −5x−6
[x3 + 5x]0 = (x3 )0 + (5x)0 = 3x2 + 5
√
√
√
[x8 − 7x]0 = (x9 )0 − ( 7x)0 = 9x8 − 7
(4x2 )0 = 4(x2 )0 = 4(2x) = 8x
= cf 0 (x)
0
0 0
x
1
1
=
(x) =
5
5
5
(log5 x)0 =
(3x7 )0 = 21x6
(3ex )0 = 3ex
1
x
1
x ln 5
10. (sin x)0 = cos x
(sin x + ex )0 = cos x + ex
11. (cos x)0 = − sin x
(5x + cos x)0 = 5 − sin x
12. (tan x)0 = sec2 x
(8x2 + tan x)0 = 16x + sec2 x
13. (sec x)0 = sec x tan x
(9x + sec x)0 = 9 + sec x tan x
14. (csc x)0 = − csc x cot x
(2x + csc x)0 = 2x ln 2 − csc x cot x
15. (cot x)0 = − csc2 x
(4ex + cot x)0 = 4ex − csc2 x
16. Product Rule: [f (x) ∗ g(x)]0 =
(x3 sin x)0 = (3x2 )(sin x) + (x3 )(cos x)
f 0 (x) ∗ g(x) + g 0(x) ∗ f (x)
0
f (x)
17. Quotient Rule:
=
g(x)
g(x) ∗ f 0 (x) − f (x) ∗ g 0 (x)
g 2 (x)
18. Inverse Rule: [f −1 (x)]0 =
provided f is differentiable and invertible and f 0 (f −1 (x)) 6= 0.
1
f 0 (f −1 (x))
x3
tan x
2
0
=
(tan x)(3x2 ) − (x3 )(sec2 x)
tan2 x
Chain Rule: If h(x) = f ◦ g(x) = f [g(x)], then h0 (x) = f 0 g(x) ∗ g 0 (x)
If h(x) = f [g(x)] we can call f (x) the Outer Function and g(x) the inner function.
Then h0 (x) = [Outer 0 (inner)] ∗ (inner)0
Equivalently, if we write y = h(x) = f (u), where u = g(x), then
dy
dy du
=
∗
dx
du dx
Dif f erentiation Rules w Chain Rule
Examples
[g(x)n ]0 = [ng(x)n−1 ] ∗ g 0 (x)
(sin3 x)0 = [(sin x)3 ]0 = 3(sin x)2 ∗ cos x = 3(sin2 x) ∗ cos x
0 √
13 0 1
2
3
3
3
5x − 7x = 5x − 7x
= 3 (5x3 − 7x)− 3 ∗ (15x2 − 7)
[ecos x ]0 = [ecos x ] ∗ (− sin x)
[eg(x) ]0 = [eg(x) ] ∗ g 0 (x)
[ln g(x)]0 =
1
∗ g 0 (x)
g(x)
1
[logb g(x)]0 =
∗ g 0 (x)
[g(x)](ln b)
[3sin x ]0 = [3sin x ln 3] ∗ (cos x)
1
0
[ln(cos x)] =
∗ (− sin x)
cos x
1
[log5 (tan x)]0 =
∗ (sec2 x)
[tan x](ln 5)
[sin g(x)]0 = [cos g(x)] ∗ g 0(x)
[sin 4x2 ]0 = [cos 4x2 ] ∗ 8x
[cos g(x)]0 = [− sin g(x)] ∗ g 0(x)
[cos(ln x)]0 = [− sin(ln x)] ∗
[tan g(x)]0 = [sec2 g(x)] ∗ g 0 (x)
[tan 9x]0 = [sec2 9x] ∗ (9)
[sec g(x)]0 = [sec g(x) tan g(x)] ∗ g 0 (x)
[sec x4 ]0 = [sec x4 tan x4 ] ∗ (4x3 )
[csc g(x)]0 = [− csc g(x) cot g(x)] ∗ g 0(x)
[csc x7 ]0 = [− csc x7 cot x7 ] ∗ (7x6 )
[cot g(x)]0 = [− csc2 g(x)] ∗ g 0(x)
[cot x5 ]0 = [− csc 2 x5 ] ∗ (5x4 )
[bg(x) ]0 = [bg(x) ln b] ∗ g 0 (x)
1
x
We use the chain rule in a recursive fashion when we have compound compositions such as
m(x) = h ◦ f ◦ g(x) = h f g(x)
m0 (x) = h0 [f (g(x))] ∗ f 0 (g(x)) ∗ g 0 (x)
Compound Chain Rule Examples:
0
[sin7 (2x3 + 4x)]0 = [sin(2x3 + 4x)]7 = 7[sin(2x3 + 4x)]6 ∗ [cos(2x3 + 4x)] ∗ (6x2 + 4)
= [7 sin6 (2x3 + 4x)] ∗ [cos(2x3 + 4x)] ∗ (6x2 + 4)
0
√
1 0
− 1
1
ecos x = ecos x 2 = ecos x 2 ∗ (ecos x ) ∗ (− sin x)
2
2
2
2
[tan(e4x )]0 = [sec2 (e4x )] ∗ (e4x ) ∗ (8x)
3