4
4.1
Differentiation
The First Principle
The above figure shows the graph of a general function y = f (x). M and N are points (a, 0)
and (a + h, 0) respectively. P and Q are the points on the curve given by x = a and x = a + h.
So M P = f (a) and N Q = f (a + h). It can be observed that
RQ = N Q − N R
= NQ − MP
= f (a + h) − f (a)
The gradient of
PQ =
=
RQ
PR
f (a + h) − f (a)
h
Hence, the gradient at
f (a + h) − f (a)
.
h
Thus, the gradient of the function f (x) at any point on the curve is given by
P = lim
h→0
f (x + h) − f (x)
.
h
f ′ (x) is called the derived function of f (x), and this method used in obtaining f ′ (x) is generally
known as the first principle.
f ′ (x) =
Example 4.1. Use the principle method to find f ′ (x) given that
(a) f (x) = x
(b) f (x) = 2x2
x
(c) f (x) =
x+1
Solution.
(a) Here, f (x) = x, f (x + h) = x + h.
f (x + h) − f (x)
(x + h) − x
h
=
= =1
h→0
h
h
h
f ′ (x) = lim
(b) f (x) = 2x2 ,
∴ f (x + h) = 2(x + h)2
= 2(x2 + 2hx + h2 )
= 2x2 + 4hx + 2h2
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
2
2x + 4hx + 2h2 − 2x2
= lim
h→0
h
4hx + 2h2
= lim
h→0
h
= lim 4x + 2h = 4x
h→0
15
(h ̸= 0).
(c) Here, f (x) =
x
x+h
, f (x + h) =
x+1
x+h+1
f ′ (x) =
=
=
=
=
=
=
=
lim
x+h
x+h+1
−
h
h→0
lim
x
x+1
(x+h)(x+1)−x(x+h+1)
(x+h+1)(x+1)
h
(x + h)(x + 1) − x(x + h + 1)
lim
h→0
h(x + h + 1)(x + 1)
x2 + hx + x + h − x2 − hx − x
lim
h→0
h(x + h + 1)(x + 1)
h
lim
h→0 h(x + h + 1)(x + 1)
1
lim
h→0 (x + h + 1)(x + 1)
1
(x + 1)(x + 1)
1
(x + 1)2
h→0
Example 4.2. Find f ′ (x) by the first principle method in each of the following:
1
(a) f (x) = x 2
(b) f (x) = xn
1
(c) f (x) = x− 2
Solution.
1
(a) Here f (x) = x 2 =
√
2. f (x + h) =
f ′ (x) =
=
=
=
=
=
√
x + h.
f (x + h) − f (x)
h→0
h
√
√
x+h− x
lim
h→0
h
√
√
√
√ x+h− x
x+h+ x
lim
×√
√
h→0
h
x+h+ x
x+h−x
lim √
√
h→0 h( x + h + x)
1
lim √
√
h→0
x+h+ x
1
1
√ = x−1/2
2
2 x
lim
(b) f (x) = xn ,
f (x + h) = (x + h)n
= xn + nxn−1 h +
n(n − 1) n−2 2
x
h + · · · + hn
2!
16
f (x + h) − f (x)
h
n
x + nxn−1 h + n(n−2)
xn−2 h2 + · · · + hn − xn
2
= lim
h→0
h
n(n−2)
nxn−1 h + 2 xn−2 h2 + · · · + hn
= lim
h→0
h
n(n − 2) n−2
n−1
n−1
+
= lim nx
x
h + ··· + h
h→0
2
f ′ (x) =
lim
h→0
= nxn−1
1
1
1
(c) Here, f (x) = x− 2 = √ and f (x + h) = √
.
x
x+h
f ′ (x) =
=
=
=
=
=
=
=
f (x + h) − f (x)
h→0
h
√1
√1
−
x
x+h
lim
h→0
h
√
√
x− x+h
lim √ √
h→0 h( x x + h)
√
√
√
√
x− x+h
x+ x+h
√
lim √ √
×√
h→0 h( x x + h)
x+ x+h
x − (x + h)
√
lim √ √
√
h→0 h( x x + h)( x + x + h)
−1
√
lim √ √
√
h→0 ( x x + h)( x + x + h)
−1
√
x×2 x
−1
lim
3
2x 2
1 3
= − x− 2
2
4.1.1
Exercise
1. Find from the first principles the derived function of the following
(a) y = x3 + 2x
(b) f (x) = x2 + x
(c) g(x) = 4x2 + 2
1
(d) h(x) = 3
2x
(e) y = x4 − x2
1
(f) f (x) =
2x + 3
1
(g) y = 3
x
2. Find the derived function of the following:
17
(a) (x2 + 1)(x − 1)
(b) (x + 21 )2
2
(c) 5
x
1
(d) √
4x2 + 1
√
x
(e) 3
x2
4.2
Techniques of Differentiation
In the preceeding examples; we used the limit definition to find derivatives. The process of
finding the derivative of a function is called differentiation. A function is differentiable at x if
its derivative exists at x, and is differentiable on an interval (a, b) if it is differentiable at every
point in the interval. In addition to f ′ (x), which is read as f prime of x. Other notations are
dy ′ d
used to denote the derivative of y = f (x). The most common are
,f,
[f (x)] and Dx [y].
dx
dx
dy
The notation
is read as the derivative of y with respect to x or simply, dy − dx. Using the
dx
limit notation, we can write
dy
f (x + h) − f (x)
= lim
= f ′ (x).
dx h→0
h
In what follows, we introduce several rules that allow us to find derivatives without the direct
use of the limit definition. We start with the simplest of all functions, the constant function
f (x) = c.
4.3
Derivatives of a Constant Function
The derivative of a constant function is 0. That is if f (x) = c, a real number, then
d
[c] = 0.
dx
For example, if y = 7, then
4.4
dy
= 0.
dx
The Power Rule
If n is any real number, then the function f (x) = xn is differentiable and
d
[f (x)] = nxn−1 .
dx
Example 4.3. Find
dy
if
dx
(a) y = x3
√
(b) y = x
Solution.
18
(a) If y = x3 , then by the power rule,
dy
= 3x3−1 = 3x2 .
dx
1
(b) We express the function in index form. That is, y = x 2 . Therefore,
dy
dx
=
=
=
4.5
1 1 −1
x2
2
1 −1
x 2
2
1
1
√
1 =
2 x
2x 2
The constant multiple rule
If c is any constant and f (x) is a differentiable function, then
d
d
[cf (x)] = c [f (x)].
dx
dx
Example 4.4. Differentiate the following
(a) y = 6x4
(b) f (x) =
4
x2
Solution.
dy
(a)
= 4(6)x4−1 = 24x3
dx
(b) f (x) can be rewritten as f (x) = 4x−2 . Thus,
f ′ (x) = −2(4)x−2−1
= −8x−3
8
= − 3
x
4.6
The Sum and Difference rule
If f and g are both differentiable, then
d
d
d
[f (x) + g(x)] =
[f (x)] +
[g(x)]
dx
dx
dx
and
d
d
d
[f (x) − g(x)] =
[f (x)] −
[g(x)].
dx
dx
dx
Example 4.5. Differentiate the following:
(a) f (x) = −2x3 + 9x2 − 2
(b) g(t) =
t4
+ 3t3 − t2 − 13t
2
Solution.
(a) f ′ (x) = −6x2 + 18x
(b) g ′ (t) =
4t3
+ 9t2 − 2t − 13
2
19
4.7
The Product Rule
If f and g are both differentiable, then
d
d
d
[f (x)g(x)] = f (x) [g(x)] + g(x) [f (x)].
dx
dx
dx
Example 4.6. Find the derivative of h(x) = (3x − 2x2 )(5 + 4x).
Solution.
d
d
[5 + 4x] + (5 + 4x) [3x − 2x2 ]
dx
dx
= (3x − 2x2 )(4) + (5 + 4x)(3 − 4x)
h′ (x) = (3x − 2x2 )
= (12x − 8x2 )(15 − 8x − 16x2 )
= −24x2 + 4x + 15
4.8
The Quotient Rule
If f and g are differentiable, then
d
d
g(x) dx
[f (x)] − f (x) dx
[g(x)]
d f (x)
=
.
2
dx g(x)
[g(x)]
Example 4.7. If y =
x2 + x − 2
dy
, find
3
x +6
dx
Solution.
dy
dx
=
=
=
=
4.8.1
d
d
(x3 + 6) dx
[x2 + x − 2] − (x2 + x − 2) dx
[x3 + 6]
(x3 + 6)2
3
(x + 6)(2x + 1) − (x2 + x − 2)(3x2 )
(x3 + 6)2
(2x4 + x3 + 12x + 6) − (3x4 + 3x3 − 6x2 )
(x3 + 6)2
−x4 − 2x3 + 6x2 + 12x + 6
(x3 + 6)2
Exercise
1. Find y ′ in each of the following
(a) y = 7x5 − 3x4 + x2
2
(b) y = 2t4 − 6t + t − 2
t
√
(c) y = 6 3 x
2. Differentiate the following functions with respect to x:
(a) f (x) = (x3 + 1)(x4 + 1)
(b) g(x) = (x + x3 )(x2 − 1)
3. Find
dy
in the following:
dx
20
x2
x+1
t3 + t
(b) y = 4
t −2
x2
√
(c) y =
1+ x
(a) y =
4.9
The Chain Rule
If y = f (u) is a differentiable function of u, and u = g(x) is a differentiable function of x, then
y = f (g(x)) is a differentiable function of x and
dy
dy
du
=
×
.
dx
du dx
Example 4.8. Differentiate the following functions:
√
(a) y = x2 + 1
(b) y = (3x2 − 4)4
Solution.
(a) Using the chain rule, we let u = x2 + 1, then y =
∴
dy
dx
√
u and
1
dy
du
= √ ,
= 2x.
du
2 u dx
dy
du
×
du dx
1
√ × 2x
2 u
x
x
√ =√
2
u
x +1
=
=
=
(b) Here again, using the chain rule, we let u = 3x2 − 4, then y = u4 and by the general power
rule,
dy
du
= 4u3 and
= 6x.
du
dx
Thus,
dy
dy
du
=
×
dx
du dx
= 4u3 × 6x
= 4(3x2 − 4)3 × 6x
= 24x(3x2 − 4)3
Example 4.9. If f (x) =
1
, find f ′ (x).
(3x − 4)2
Solution. Note that you don’t need to use the quotient rule every time you see a quotient. Here
we can rewrite the function as
f (x) = (3x − 4)−2 .
If we let u = 3x − 4, then f (u) = u−2 and by the general power rule,
f ′ (u) = −2u−3 and u′ (x) = 3.
21
Thus,
f ′ (x) = f ′ (u) × u′ (x)
= −2u−3 × 3
= −6u−3
6
= − 3
u
−6
=
(3x − 4)3
4.10
Derivatives of Trigonometric Functions
We now extend our technique of differentiation to include the trigonometric functions. We begin
by stating a special theorem on trigonometric limits without proof, which shall be required later.
Theorem 4.10.
sin x
1 − cos x
= 1 and lim
=0
x→0 x
x→0
x
lim
This result plays a crucial part in finding the derivatives of sin x and cos x. To obtain their
derivatives, we must work in radians.
4.11
Derivative of sin x and cos x
If f (x) = sin x, then
f ′ (x) =
=
=
=
=
=
=
=
=
=
f (x + h) − f (x)
h→0
h
sin(x + h) − sin x
lim
h→0
h
sin x cos h + cos x sin h − sin x
lim
h→0
h
sin x cos h − sin x + cos x sin h
lim
h→0
h
sin x(cos h − 1) + cos x sin h
lim
h→0
h cos h − 1
sin h
lim sin x
+ cos x
h→0
h
h
cos h − 1
sin h
lim sin x
+ lim cos x
h→0
h→0
h
h
cos h − 1
sin h
+ cos x lim
sin x lim
h→0
h→0
h
h
1 − cos h
sin h
− sin x lim
+ cos x lim
h→0
h→0
h
h
− sin x(0) + cos x(1)
lim
= cos x
Thus,
d
[sin x] = cos x.
dx
22
Similarly, if f (x) = cos x, then f ′ (x) = − sin x. That is,
d
[cos x] = − sin x. (Prove as an exercise).
dx
Example 4.11. Differentiate the following functions:
(a) y = sin 3x
cos x
(b) f (x) =
2
(c) g(x) = sin(4x2 − 3)
(d) y = x6 cos x + 2x5
Solution.
(a) Here, we let u = 3x, then y = sin u by the chain rule. Thus,
dy
dx
dy
du
×
du dx
= cos u × 3
=
= 3 cos u
= 3 cos 3x
(b) Let f (x) =
cos x
1
= cos x. Thus,
2
2
f ′ (x) =
1
1
× − sin x = − sin x.
2
2
(c) Let u = 4x2 − 3, then g(u) = sin u and by the chain rule,
g ′ (x) = g ′ (u) × u′ (x)
= − cos u × 8x
= −8x × cos u
= −8x cos(4x2 − 3)
(d)
dy
dx
d
d
d
[cos x] + cos x [x6 ] +
[2x5 ]
dx
dx
dx
= x6 × (− sin x) + cos x × 6x5 + 10x4
= x6
= −x6 sin x + 6x5 cos x + 10x4
= x4 (−x2 sin x + 6x cos x + 10)
Example 4.12. Differentiate the following with respect to x.
(a) tan x
(b) cosec x
(c) sin3 x
1 − cos x 2
(d)
sin x
Solution.
23