Derivatives of Polynomial Functions
The derivative of a polynomial is one of the fundamental applications of differential calculus and holds a prominent place in mathematics. The derivative of a polynomial function is evaluated by differentiating each term individually according to the linearity of the derivative operator.
This topic will be explored comprehensively under the core calculus curriculum.
Let us evaluate the successive derivatives of the monomial given by \( P(x) = ax^n \).
\[P(x) = ax^n \]
\[ P'(x) = a \cdot n x^{n-1} \quad (\text{where } P'(x) \text{ denotes the first derivative of } P(x).) \]
\[ P^{”}(x) = a \cdot n \cdot (n – 1) x^{n-2} \quad (\text{where } P^{”}(x) \text{ denotes the second derivative of } P(x).) \]
\[ P^{”’}(x) = a \cdot n \cdot (n – 1) \cdot (n – 2)x^{n-3}
\quad (\text{where } P^{”’}(x) \text{ denotes the third derivative of } P(x).)\]
\[ \vdots\]
\[P^{(m)}(x) = a \cdot n \cdot (n – 1) \cdot (n – 2) \dots (n – m + 1) x^{n-m}
\quad (\text{where } P^{(m)}(x) \text{ denotes the } m\text{-th derivative of } P(x).)\]
\[ \text{The derivative of a constant polynomial } P(x) = a_0 \text{ is given by } P'(x) = 0. \]
To differentiate a polynomial $P(x)$ consisting of multiple terms, the derivative of each term is computed independently.
Example:
\[ \text{Given } P(x) = 2x^3 + 5x^2 – x + 2, \]
\[ P'(x) = 2 \cdot 3x^{3-1} + 5 \cdot 2x^{2-1} – 1x^{1-1} + 0 = 6x^2 + 10x – 1 \]
\[ P^{”}(x) = 6 \cdot 2x^{2-1} + 10 \cdot 1x^{1-1} – 0 = 12x + 10 \]
\[ P^{”’}(x) = 12 \cdot 1x^{1-1} + 0 = 12 \]
4. If a polynomial \( P(x) \) is exactly divisible by \( (ax + b)^2 \),
then \( P'(x) \) is also exactly divisible by \( ax + b \). In this case,
\[
P\left(- \frac{b}{a} \right) = 0 \quad \text{and} \quad P’\left(- \frac{b}{a} \right) = 0
\]
Similarly, if the polynomial \( P(x) \) is exactly divisible by \( (ax + b)^3 \), then both \( P'(x) \) and \( P^{”}(x) \) are exactly divisible by \( ax + b \). Consequently,
\[
P\left(- \frac{b}{a} \right) = 0, \quad P’\left( -\frac{b}{a} \right) = 0, \quad \text{and} \quad P^{”}\left( -\frac{b}{a} \right) = 0
\]
Generalizing this principle: If a polynomial \( P(x) \) is exactly divisible by \( (ax + b)^n \), then the successive derivatives \( P'(x) \), \( P^{”}(x) \), \( P^{”’}(x) \), \( \dots \), \( P^{(n-1)}(x) \) are all exactly divisible by \( ax + b \). Thus,
\[
P\left(- \frac{b}{a} \right) = 0,
\]
\[
P’\left( -\frac{b}{a} \right) = 0,
\]
\[
P^{”}\left(- \frac{b}{a} \right) = 0,
\]
\[
\vdots
\]
\[
P^{(n-1)}\left(- \frac{b}{a} \right) = 0
\]
Example:
\[
P(x) = 8x^3 + 4ax^2 – 2bx + 1
\]
Assuming the polynomial is exactly divisible by \( (2x – 1)^2 \), let us determine the value of \( a \).
\[
P\left( \frac{1}{2} \right) = 8 \cdot \left(\frac{1}{2}\right)^3 + 4a \cdot \left(\frac{1}{2}\right)^2 – 2b \cdot \left(\frac{1}{2}\right) + 1 = 0
\]
\[
\Rightarrow a – b = -2
\]
\[
P'(x) = 24x^2 + 8ax – 2b
\]
\[ P’ \left( \frac{1}{2} \right) = 24 \cdot \frac{1}{2^2} + 8a \cdot \frac{1}{2} – 2b = 0 \]
\[ \Rightarrow 2a – b = -3 \]
Solving this system of simultaneous linear equations yields:
\[ a = -1 \]
Example:
If the polynomial \( P(x) = x^5 – ax^3 + bx^2 + x + c \) is exactly divisible by \( (x – 1)^3 \), let us determine the value of \( c \).
\[ P(1) = 1 – a + b + 1 + c = 0 \Rightarrow -a + b + c = -2\]
\[ P'(x) = 5x^4 – 3ax^2 + 2bx + 1 \]
\[ P'(1) = 5 – 3a + 2b + 1 = 0 \Rightarrow -3a + 2b = -6 \]
\[ P^{”}(x) = 20x^3 – 6ax + 2b \]
\[ P^{”}(1) = 20 – 6a + 2b = 0 \Rightarrow -3a + b = -10 \]
Solving this system of simultaneous linear equations yields:
\[ c = -\frac{4}{3} \]
Example:
Given that the remainder of the polynomial \( P(x) = x^5 + ax^3 + 2x^2 – x + b \) when divided by \( (x – 1)^2 \) is \( x + 5 \), let us find the values of \( a \) and \( b \).
Since the division of \( P(x) \) by \( (x – 1)^2 \) leaves a remainder of \( x + 5 \), the modified polynomial
\[ R(x) = P(x) – (x + 5) = x^5 + ax^3 + 2x^2 – 2x + b – 5 \]
must be exactly divisible by \( (x – 1)^2 \). Therefore,
\[ R(1) = 1^5 + a \cdot 1^3 + 2 \cdot 1^2 – 2 \cdot 1 + b – 5 = 0 \Rightarrow a + b = 4 \]
\[ R'(x) = 5x^4 + 3ax^2 + 4x – 2 \]
\[ R'(1) = 5 \cdot 1^4 + 3a \cdot 1^2 + 4 \cdot 1 – 2 = 0 \Rightarrow a = -\frac{7}{3} \]
Substituting \( a \) back into the linear relation
\[ a + b = 4 \]
yields:
\[ b = \frac{19}{3} \]