These notes explain in an easily understandable manner, every single rule and technique in order to master the process of differentiating mathematical functions.
To a certain extent, Precalculus isn't even needed to comprehend these notes, as every single method is fully explained, and no-nonsen...
,Part I
Differentiation
1 Simple Rules of Differentiation
Differentiation is the process of finding a derivative or gradient function.
Given a function f (x) we obtain f 0 (x) by differentiating with respect to the variable x.
There are a number of rules associated with differentiation. These rules can be used to
differentiate more complicated functions without having to use first principles.
We can summarize the following rules:
f (x) f 0 (x) Name of Rule
c (a constant) 0 Differentiating a Constant
xn nxn−1 Differentianting xn / Power Rule
cu(x) cu0 (x) Constant times a function
u(x) + v(x) u0 (x) + v 0 (x) Addition Rule
The last two rules can be proved with the first principles of differentiation:
4
,2 The Power Rule
2.1 The Power Rule
We start with the derivative of a power function, f (x) = xn . Here n is a number of
any kind: integer, rational, positive, negative, even irrational, as in xπ . The formula to
differentiate xn :
d n
x = nxn−1
dx
2.2 Proof of The Power Rule
We can prove this with the definition of the derivative and the binomial expansion.
Find the derivative of f (x) = xn :
First, we know the definition of the derivative is:
f (x + h) − f (x)
lim
h→0 h
We apply this to xn :
(x + h)n − xn
lim
h→0 h
We know that the binomial expansion for (x + h)n is:
n
n
X n n−r r
(x + h) = x h
r=0
r
n
X n n−r r n n n n−1 n n−2 2 n n
x h = x + x h+ x h + ... + h
r=0
r 0 1 2 n
We introduce this expansion into the limit:
n
x + n1 xn−1 h + n2 xn−2 h2 + ... + n
n n
0 n
h − xn
lim
h→0 h
We cancel out terms:
n n n n
n
n−1 n−2 2 n
n
−x
lim 0x + 1
x h+ 2
x h + ... + n
h
h→0 h
n n n
1
xn−1 h + 2
xn−2 h2 + ... + n
hn
lim
h→0 h
We then cancel h:
h hn−1
n n n
n−1
2
n−2
>
n
1
x h+
2
x h + ... + n
h
lim
h→0 h
n n−1 n n−2 n n−1
lim x + x h + ... + h
h→0 1 2 n
5
, We could then factorize:
!
n n−1 n n−2 n n−1
lim x +h x + ... + 1
h→0 1 2 n
We can then evaluate the limit:
:!0
n n−1 n n−2 n n−1
n n−1
lim x +h x +
... + 1 = x
h→0 1 2 n 1
∴ f 0 (x) = nxn−1
2.3 Solved Exercises
1. Find the derivatives of the given functions.
(a) x100 (d) xπ
3
(b) x−100 (e) x 4
1 √
(c) x5
(f) 3 x
(a)
d 100
x = 100x99
dx
(b)
d −100
x = −100x−101
dx
(c)
d 1 d −5
≡ x = −5x−6
dx x5 dx
(d)
d π
x = πxπ−1
dx
(e)
d 3 3 3 3 5
x 4 = x( 4 −1) = x− 4
dx 4 4
(f)
d √ d 1 3 1
3 x≡ 3x 2 = x− 2
dx dx 2
6
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