# Derivatives of Polynomials

```Lesson – 2
Introduction to Calculus.
The Derivative
of the
polynomial function
One of main problem that calculation deals with is finding of slope
of the tangent line at a point on a curve. Let consider y = f(x) on figure 2.
1.
Fig. 2.1.
We can see on figure above, points A and B have the co-ordinates
[x1, f (x1)] and [x1 + x, f (x1 + x)], respectively.
The line passing through points A ad B is called a secant line. The
slope of this line is given well-known formula.
f ( x2 )  f ( x2 )
m = f ( x ) =
x
x2  x1
(2.1)
Suppose we keep point A fixed in position but move only point B
along a curve so x approaches point A. As point B gets closer and
closer to point A, the secant line a limiting position at point A and this line
are called the tangent line.
Now, since the tangent line is a limiting position of secant lines, we
consider the slope of the tangent line to a limiting value of the slopes of
secant lines A-B as point B approaches point A. This occurs when x
approaches 0 and is seen in Fig. 2. 2.
Fig. 2. 2.
We may say that the slope of a curve y = f(x) at point [x1, f(x1)]
point A) is given by the limit:
m=
lim
x  0
f ( x )
=
x
lim
x  0
f ( x1  x)  f ( x1 )
x
We can make this result apply to any points [x, f (x)] on curves just
by replacing x1 with x. These give us:
lim
x  0
f ( x1  x)  f ( x1 )
x
(2. 2)
This limit is basis of the differential calculus and is called the
derivative of f (x) with respect to x and is denoted as f’ of x, or called
d [ f ( x)]
dx
(dee f(x), dee x) or dy (dee y, dee x)
dx
This process of finding derivatives is called differentiations.
Obtaining derivatives of functions using the formula, we called obtaining
derivatives from the first principle.
Remember:
dy y
but

dx x
y
lim x
x  0
Example 1.
If f (x) = x2 – 2x, find derivative f’ (x) from the first principle.
Solution:
For the function f (x) = x2 – 2x the f (x + x) is determined as:
f (x) = x2 – 2x
then difference quotient is equal to:
2
2
y
= f ( x  x)  f ( x) = ( x  x)  2( x  x)  ( x  2 x)
x
x
x
as (x+ x)2 = x2 + 2xx + x
(from known formula (m + n)2 = m2 + 2mn + n2
then:
2
2
2
2
y
= x  2 xx  (x)  2 x  2 x  2x  x  2 x = 2 xx  x  2x
x
x
x
After factoring x from nominator and cancelling, we obtain:
y
= x(2 x  x  2) = 2x + x – 2
x
x
To get the derivative, we take the limits as x → 0. To doing
this we, treat x as a constant because it is x that is changing.
f’ (x) = dy =
dx
lim (2 x  x  2)
= 2x – 2
x  0
Example 2.
If y = 4 , find dy from the first principle.
x
dx
Solution:
If y = f x) = 4 , then difference quotient is equal to:
x
4 x  ( x  x)4
4
4

y
= f ( x  x)  f ( x) = x  x x = x( x  x) =
x
x
x
x
= 4 x  4 x  4 x =
x ( x  x )  x
 4x
4
4
=
=x( x  x)x
x( x  x)
x( x  x)
Thus,
F’(x) = dy =
dx
lim
4
x( x  x)
= - 42
x
x  0
Rules for differentiation to Remember
1.
If f (x) = C then f’ (x) = 0
2.
If f (x) = xn then f’ (x) = nxn – 1
3.
If g (x) = C*f (x), g’(x) = C*f’(x)
4.
If f (x) = g (x) &plusmn; h (x) then
f’(x) = g’(x) &plusmn; h’(x)
Example 3.
Find datives following functions:
a)
y = x4
Solution:
y’= 4x3
b)
y = 5/4x(5/4 – 1)
Solution:
y’ = 5/4(5/4-1)x(5/4-1 – 1) = 5/4(1/4)x(5/4- 2) = 5/16x-3/4
c)
y = x-6
Solution:
y’ = -6x(-6 – 1) = -6x-7
d)
s = 31
t
Solution:
We must write s as a power of t. So
s = t-1/3
s’ = -1/3 t-1/3- 1 = -1/3t-4/3 = -1/3t-4/3 = -1/3 1 = -1/3 31 =
3
t4
t t
=-1/3t-4/3
e)
y = -6t-3
Solution:
y’= -6(-3)t(-3 – 1) = 18t-4
f)
1
= 18 t 4 = 184
t
y = -6x3 – 2x2 + 7x – 8
Solution:
y’= -6(3)x2 –2(2)x + 7x1 – 1 –0 = -18x2-4x + 7x0 = -18x2 –4x + 7
g)
f (h) = 5h(h2 – 6), find f’ (2)
Solution:
f’ (h) = 15h2 – 30h
and
f’ (2) = 60 – 30 = 30
EXERCISES:
Differentiate following functions:
1.
f (x) = 5
2.
f (x) = 26
3.
y = x6
4.
f (x) = x21
5.
y = x80
6.
y = x6.1
7.
f (x) = 9x2
8.
y = 4x4
9.
g (w) = 4w2
10.
v (x) = 2x3
11.
y = 2/3x4
12.
f (p) = 3 p4
13.
9
f (t) = t
14.
3
y= x
15.
f (x) = x + 3
16.
f (x) = 3x – 2
18
3
17.
f (x) = 3x2 –2x + 3
18.
f (x) = 7x2 – 5x
19.
g (p) = p4 – 3p3 – 1
20.
f (t) = -13t2 + 14t + 1
21.
y = x-8 + x5
22.
y = -8x4 + 6
23.
y = -13x3 + 14x2 –2x + 3
24.
V (r) = r8 –r6 + 3r2 + 1
25.
h (x) = 4x4 + x3 –9/2x2 +9x
26.
f (x) = -3x2 + 9/2x + 2
27.
f(x) = 3/2x4 7/3x3
28.
p (x) = x7/7 + 2/3x
29.
f (x) = x7/2
30.
f (x) = 2x-14/5
31.
y = x3/4
32.
y = -9x1/3 + 5x-2/5
33.
y=x
34.
y = -x-2/3
35.
f (r) = 6 3 r
36.
8
y = 4x x
37.
f (x) = x-4
38.
f (s) = 3s-2
39.
f (x) = x-3 + x-5 – 2x-6
40.
f (x) = 100x-3 + 10x1/2
41.
y = 17
42.
y = 23
43.
y = 15
44.
y = 15
x
x
x
4x
1.
0
6.
2.
6x5
0
3.
6.1x5.1
7.
18x
10.
6x2
11.
8/3x3
14.
x2
15.
1
16.
19.
4p3 – 9p4
20.
-26t + 14
23.
-39x2 + 28x – 2 24.
26.
-6x + 9/2
30.
4.
21x20
5.
80x79
8.
16x3
9.
20w4
12.
4 3 p3
3
17.
6x – 2
21.
-8x-9 + 5x4 22.
8r7 –42r5 + 6r
13.
25.
1/2t8
18.
28.
x6 + 2/3
-28/5x-19/5 31.
3/4x-1/4
32.
-3x-2/3 –2x-7/5
33.
1/2x-1/2
34.
2/3x-1/3
35.
2r-2/3
37.
-4x-5
38.
-6s-3 39.
-3x-4 – 5x-6 + 12x-7
40.
-300x-4 + 5x-1/2
41.
-7x-8
42.
-32x3
16x3 + 3x2 9x + 9
6x3 + 7x2
27.
14x – 5
29.
36.
6x-4 43.
7/2x5/2
1/2x-7/8
-15x-6
44. -5/4x-6
```