In the previous section,
we saw the algebra of derivatives. We also saw the derivative f'(x) when f(x) = x2. In this section, we will see the derivative when the power of x is any +ve integer. Later in this section, we will see derivatives of polynomial functions also.
Derivative when the power of x is any positive integer
It can be written in 4 steps:
1. We want f'(x) where f(x) = xn
2. We can write:
f′(x) = limh→0[f(x+h)–f(x)h] = limh→0[(x+h)n–xnh]
3. So first we need to find (x+h)n–xnh
• It can be calculated as follows:
(x+h)n = (nC0)xn + (nC1)xn−1h + (nC2)xn−2h2 + (nC3)xn−3h3 + . . . + (nCn)hn - - - (I)⇒(x+h)n = xn + nxn−1h + (nC2)xn−2h2 + (nC3)xn−3h3 + . . . + nhn⇒(x+h)n − xn = nxn−1h + (nC2)xn−2h2 + (nC3)xn−3h3 + . . . + nhn⇒(x+h)n − xn = h[nxn−1 + (nC2)xn−2h1 + (nC3)xn−3h2 + . . . + nhn−1]⇒(x+h)n − xnh = nxn−1 + (nC2)xn−2h1 + (nC3)xn−3h2 + . . . + nhn−1
◼ Remarks:
•
Line marked as I:
In this line, we apply the binomial theorem.
4. Now the calculations in (2) can be completed:
f′(x) = limh→0[f(x+h)–f(x)h] = limh→0[(x+h)n–xnh] = limh→0[nxn−1 + (nC2)xn−2h1 + (nC3)xn−3h2 + . . . + nhn−1] = nxn−1 - - - (I)
◼ Remarks:
•
Line marked as I:
When the limit is applied, all terms with 'h' will become zero.
This formula can be proved using mathematical induction also. It can be written in 5 steps:
1. We want to prove the statement:
If f(x) = xn, then f'(x) = nxn-1.
2. First we must prove the statement for n = 1
♦ If n = 1, then f(x) = x1 = x
♦ Then f'(x) = 1 × x1-1 = 1 × x0 = 1
• So the statement is true for n = 1.
3. Next we assume that the statement is true for n = k
♦ If n = k, then f(x) = xk
♦ Then f'(x) = k × xk-1
•
That is., we assume that, f'(xk) = k × xk-1 is true.
4. Next we must prove that, the statement is true for n = k+1
♦ If n = k+1, then f(x) = xk+1
♦ Then f'(x) = (k+1) × x(k+1)-1 = (k+1) × xk .
•
That is.,we must prove that,
f'(xk+1) = (k+1) × x(k+1)-1 = (k+1) × xk is true.
5. This can be done as follows:
f′(xk+1) = (k+1)xk⇒f′(xk.x) = (k+1)xk⇒f′(xk).x + xk.f′(x) = (k+1)xk - - - (I)⇒f′(xk).x + xk.1 = (k+1)xk - - - (II)⇒k.xk−1.x + xk.1 = (k+1)xk - - - (III)⇒k.xk + xk = (k+1)xk⇒(k+1)xk = (k+1)xk
◼ Remarks:
• Line marked as I:
Here we use the product rule on the LHS
• Line marked as II:
Here we use the fact that:
If f(x) = x, then f'(x) = 1
• Line marked as III:
Here we use the result in (3)
•
So now we have a useful formula:
If f(x) = xn, then f'(x) = nxn-1.
♦ We proved this when n is a +ve integer.
♦ In fact, this formula is valid when n is any real number. We will see it's proof in higher classes.
•
This formula is known as Theorem 6.
Let us see an example.
If f(x) = x9, then f'(x) will be 9x8
◼ A graphical description can be written in 5 steps:
1. We know that, f’(x) is a function. So we must be able to plot f’(x).
2. In fig.13.38 below, both f(x) and f’(x) are plotted in the same graph.
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| Fig.13.38 |
3. The two graphs are plotted in different colors.
♦ The red curve represents f(x)
♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at (x = 0.75). That is., we want f’(0.75)
• For that, we mark point A on the red curve. Here, (x = 0.75)
• Next we draw a vertical line through A. This vertical line meets the green curve at A'.
• f’(0.75) will be equal to the y-coordinate of A'. In this case, it is 0.9.
•
We can verify this theoretically:
f'(0.75) = 9 × (0.75)8 = 0.9010
(ii) Suppose that, we want the derivative of f(x) at (x = -0.8). That is., we want f’(-0.8)
• For that, we mark point B on the red curve. Here, (x = -0.8)
• Next we draw a vertical line through B. This vertical line meets the green curve at B'.
• f’(-0.8) will be equal to the y-coordinate of B'. In this case, it is 1.51.
•
We can verify this theoretically:
f'(-0.8) = 9 × (-0.8)8 = 1.509
5. Further more, the reader may verify the tangents also:
•
Through the point A, draw a line at a slope of 0.9. This line will be tangential to f(x).
•
Through the point B, draw a line at a slope of 1.51. This line will be tangential to f(x).
Derivative of polynomial functions
This can be written in 5 steps:
1. Consider the polynomial function:
f(x) = anxn + an-1 xn-1 + an-2 xn-2 + . . . + a1x1 + a0 x0.
♦ a0, a1, a2 etc., are real numbers.
•
We want to find f'(x)
2. We can consider each term as a separate function.
•
Then we can find the derivative of each function using theorem 6 that we saw above.
•
For example, the derivative of an-1 xn-1 will be an-1 (n-1)xn-2.
3. After finding each derivative, we can apply the sum rule.
4. So we get:
f'(x) = an nxn-1 + an-1 (n-1)xn-2 + an-2 (n-2)xn-3 + . . . + a1(1)x0 + 0.
5. This method for finding the derivative of polynomial functions is known as theorem 7.
Let us see some solved examples:
Solved example 13.14
Compute the derivative of 6x100 - x55 + x
Solution:
1. Derivative of the first term is:
6 × 100x99 = 600x99.
2. Derivative of the second term is:
- 55x54.
3. Derivative of the third term is: 1.
4. Applying the sum rule, we get:
f'(x) = 600x99 - 55x54 + 1
Solved example 13.15
Compute the derivative of
1 + x + x2 + x3 + . . . + x50 at x = 1
Solution:
1. We see that, there are 51 terms in f(x).
2. The derivative f'(x) is:
0 + 1 + 2x + 3x2 + 4x3 + . . . + 50x49
•
So there are 50 terms in f'(x)
3. Substituting (x=1) in f'(x), we get:
f'(1) = 1 + 2 + 3 + 4 + . . . + 50
•
This is in the form of an A.P where,
♦ First term = 1
♦ Last term = 50
♦ Number of terms (n) = 50
4. So sum of the A.P can be obtained as:
S = n(First term + Last term)2 = 50(1+50)2 = 25 × 51 = 1275
Solved example 13.16
Find the derivative of f(x)=x+1x
Solution:
1. We can apply the quotient rule:
(uv)′=u′v−uv′v2
2. In our present case:
♦ u = (x+1). So u' = 1
♦ v = x. So v' = 1
3. Thus we get:
(uv)′=1(x)−(x+1)1x2=x−x−1x2=−1x2
In the next section, we will see derivatives of trigonometric functions.
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