Chapter 13.16 - Derivatives of Polynomial Functions

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:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(x+h)^n – x^n}{h} \right]}}
&{} \\

\end{array}$

3. So first we need to find $\frac{(x+h)^n – x^n}{h}$
• It can be calculated as follows:

$\begin{array}{ll}
{}&{(x+h)^n}
& {~=~}& {\rm{\left(nC_0 \right)}x^n~+~\rm{\left(nC_1 \right)}x^{n-1} h~+~\rm{\left(nC_2 \right)}x^{n-2} h^2~+~\rm{\left(nC_3 \right)}x^{n-3} h^3~+~.~.~.~+~\rm{\left(nC_n \right)}h^n ~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{(x+h)^n}
& {~=~}& {x^n~+~\rm{n}x^{n-1} h~+~\rm{\left(nC_2 \right)}x^{n-2} h^2~+~\rm{\left(nC_3 \right)}x^{n-3} h^3~+~.~.~.~+~\rm{n}h^n}
&{} \\

{\Rightarrow}&{(x+h)^n~-~x^n}
& {~=~}& {\rm{n}x^{n-1} h~+~\rm{\left(nC_2 \right)}x^{n-2} h^2~+~\rm{\left(nC_3 \right)}x^{n-3} h^3~+~.~.~.~+~\rm{n}h^n}
&{} \\

{\Rightarrow}&{(x+h)^n~-~x^n}
& {~=~}& {h\left[\rm{n}x^{n-1}~+~\rm{\left(nC_2 \right)}x^{n-2} h^1~+~\rm{\left(nC_3 \right)}x^{n-3} h^2~+~.~.~.~+~\rm{n}h^{n-1}\right]}
&{} \\

{\Rightarrow}&{\frac{(x+h)^n~-~x^n}{h}}
& {~=~}& {\rm{n}x^{n-1}~+~\rm{\left(nC_2 \right)}x^{n-2} h^1~+~\rm{\left(nC_3 \right)}x^{n-3} h^2~+~.~.~.~+~\rm{n}h^{n-1}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we apply the binomial theorem.

4. Now the calculations in (2) can be completed:

$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(x+h)^n – x^n}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\rm{n}x^{n-1}~+~\rm{\left(nC_2 \right)}x^{n-2} h^1~+~\rm{\left(nC_3 \right)}x^{n-3} h^2~+~.~.~.~+~\rm{n}h^{n-1} \right]}}
&{} \\

{}&{}
& {~=~}& {{n x^{n-1}} ~ \color {green} {\text{- - - (I)}}}
&{} \\

\end{array}$

◼ 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) = nx^n-1.
2. First we must prove the statement for n = 1
   ♦ If n = 1, then f(x) = x¹ = x
   ♦ Then f'(x) = 1 × x^1-1 = 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) = x^k
   ♦ Then f'(x) = k × x^k-1
• That is., we assume that, f'(x^k) = k × x^k-1 is true.
4. Next we must prove that, the statement is true for n = k+1
   ♦ If n = k+1, then f(x) = x^k+1
   ♦ Then f'(x) = (k+1) × x^(k+1)-1 = (k+1) × x^k .
• That is.,we must prove that,
f'(x^k+1) = (k+1) × x^(k+1)-1 = (k+1) × x^k is true.
5. This can be done as follows:

$\begin{array}{ll}
{}&{f'(x^{k+1})}
& {~=~}& {(k+1)x^k}
&{} \\

{\Rightarrow}&{f'(x^k \; .\; x)}
& {~=~}& {(k+1)x^k}
&{} \\

{\Rightarrow}&{f'(x^k) \; .\; x ~+~x^k \; .\; f'(x)}
& {~=~}& {(k+1)x^k ~~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{f'(x^k) \; .\; x ~+~x^k \; .\; 1}
& {~=~}& {(k+1)x^k ~~ \color {green} {\text{- - - (II)}}}
&{} \\

{\Rightarrow}&{k\; .\; x^{k-1} \; .\; x ~+~x^k \; .\; 1}
& {~=~}& {(k+1)x^k ~~ \color {green} {\text{- - - (III)}}}
&{} \\

{\Rightarrow}&{k\; .\; x^{k}~+~x^k}
& {~=~}& {(k+1)x^k}
&{} \\

{\Rightarrow}&{(k+1)x^k}
& {~=~}& {(k+1)x^k}
&{} \\

\end{array}$

◼ 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)

◼ Thus we have proved the statement in (4). And so, the statement in (1) is proved.

---

• So now we have a useful formula:
If f(x) = xⁿ, then f'(x) = nx^n-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.

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) = a_nxⁿ + a_n-1 x^n-1 + a_n-2 x^n-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 a_n-1 x^n-1 will be a_n-1 (n-1)x^n-2.
3. After finding each derivative, we can apply the sum rule.
4. So we get:
f'(x) = a_n nx^n-1 + a_n-1 (n-1)x^n-2 + a_n-2 (n-2)x^n-3 + . . . + a₁(1)x⁰ + 0.
5. This method for finding the derivative of polynomial functions is known as theorem 7.

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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:
$S~=~\frac{n(\text{First term + Last term})}{2}~=~\frac{50(1+50)}{2}$ = 25 × 51 = 1275

Solved example 13.16
Find the derivative of $f(x)=\frac{x+1}{x}$
Solution:
1. We can apply the quotient rule:
$\left(\frac{u}{v} \right)' = \frac{u'v - uv' }{v^2}$
2. In our present case:
    ♦ u = (x+1). So u' = 1
    ♦ v = x. So v' = 1
3. Thus we get:
$\left(\frac{u}{v} \right)' = \frac{1(x) - (x+1)1 }{x^2} = \frac{x - x - 1}{x^2} = - \frac{1}{x^2}$

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In the next section, we will see derivatives of trigonometric functions.

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