Derivative of Algebraic Functions

Sahabat Smaridasa - Mathematical Concepts

Previously, we studied the "General Definition of Derivative", where the derivative of a function \(f(x)\), denoted \(f'(x)\) or \(y'\), is defined using the limit:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

However, solving limits for algebraic functions using this definition can be lengthy and complicated. To simplify, we will discuss Derivatives of Algebraic Functions using basic derivative rules derived from the general definition.

Basic Derivative Rules for Algebraic Functions

• i) \(y = k \quad \Rightarrow \quad y' = 0\), where \(k\) is a constant.
• ii) \(y = ax^n \quad \Rightarrow \quad y' = n \cdot a \cdot x^{n-1}\), where \(n\) is any real number.
• iii) \(y = U \pm V \quad \Rightarrow \quad y' = U' \pm V'\)
• iv) \(y = U \cdot V \quad \Rightarrow \quad y' = U' \cdot V + U \cdot V'\) (Product Rule)
• v) \(y = \frac{U}{V} \quad \Rightarrow \quad y' = \frac{U' \cdot V - U \cdot V'}{V^2}\) (Quotient Rule)
• vi) \(y = [g(x)]^n \quad \Rightarrow \quad y' = n \cdot [g(x)]^{n-1} \cdot g'(x)\) (Chain Rule)
• vii) \(y = f[g(x)] \quad \Rightarrow \quad y' = f'[g(x)] \cdot g'(x)\) (Composite Function)

📘 Example 1 : Basic Power Rule

Find the derivatives of the following functions:

a) \(y = 3\)    b) \(y = x^5\)    c) \(y = \frac{5}{x^2}\)    d) \(y = 3\sqrt{x}\)    e) \(y = \frac{2}{3x\sqrt{x}}\)

Solution:

a) Using rule i: \(y' = 0\).

b) Using rule ii with \(n = 5\): \(y' = 5x^{4}\).

c) \(y = 5x^{-2} \Rightarrow y' = -10x^{-3} = -\frac{10}{x^3}\).

d) \(y = 3x^{1/2} \Rightarrow y' = \frac{3}{2}x^{-1/2} = \frac{3}{2\sqrt{x}}\).

e) \(y = \frac{2}{3}x^{-3/2} \Rightarrow y' = \frac{2}{3} \cdot (-\frac{3}{2})x^{-5/2} = -x^{-5/2} = -\frac{1}{x^2\sqrt{x}}\).

📘 Example 2 : Sum/Difference Rule

Find \(f'(x)\) for:

a) \(f(x) = 3x^2 - 2x\)    b) \(f(x) = 2\sqrt{x} + 5x^3 - 7\)    c) \(f(x) = x^5 + 2x^3 - 3x + 1\)

Solution:

a) \(f'(x) = 6x - 2\).

b) \(f'(x) = \frac{1}{\sqrt{x}} + 15x^2\).

c) \(f'(x) = 5x^4 + 6x^2 - 3\).

📘 Example 3 : Product Rule

Find the derivative of \(y = (x^2 - 1)(2x^3 + x)\).

Solution: Let \(U = x^2 - 1\), \(V = 2x^3 + x\). Then \(U' = 2x\), \(V' = 6x^2 + 1\).

\[ y' = U'V + UV' = 2x(2x^3 + x) + (x^2 - 1)(6x^2 + 1) \] \[ = 4x^4 + 2x^2 + (6x^4 + x^2 - 6x^2 - 1) = 10x^4 - 3x^2 - 1. \]

📘 Example 4 : Quotient Rule

Find the derivative of \(y = \frac{x^2 + 2}{3x - 5}\).

Solution: Let \(U = x^2 + 2\), \(V = 3x - 5\). Then \(U' = 2x\), \(V' = 3\).

\[ y' = \frac{U'V - UV'}{V^2} = \frac{2x(3x-5) - (x^2+2)(3)}{(3x-5)^2} = \frac{6x^2 - 10x - 3x^2 - 6}{9x^2 - 30x + 25} = \frac{3x^2 - 10x - 6}{9x^2 - 30x + 25}. \]

📘 Example 5 : Chain Rule (Power of a Function)

Find the derivative of \(y = (2x^2 - 3x + 8)^{10}\).

Solution: Let \(g(x) = 2x^2 - 3x + 8\), so \(g'(x) = 4x - 3\).

\[ y' = 10(2x^2 - 3x + 8)^9 \cdot (4x - 3) = (40x - 30)(2x^2 - 3x + 8)^9. \]

📘 Example 6 : Composite Function (Chain Rule)

Given \(f(2x - 1) = 3x^2 + 2x + 5\), find \(f'(3)\).

Solution: Differentiate both sides with respect to \(x\):

Left side derivative using chain rule: \(2f'(2x - 1)\). Right side derivative: \(6x + 2\).

So \(2f'(2x - 1) = 6x + 2 \Rightarrow f'(2x - 1) = 3x + 1\).

To find \(f'(3)\), set \(2x - 1 = 3 \Rightarrow x = 2\). Then \(f'(3) = 3(2) + 1 = 7\).

📘 Example 7 : Evaluating Derivatives at a Point

Find \(f'(1)\) for each function from previous examples.

Solution:

a) \(f(x) = x^5 \Rightarrow f'(x) = 5x^4 \Rightarrow f'(1) = 5\).

b) \(f(x) = 2\sqrt{x} + 5x^3 - 7 \Rightarrow f'(x) = \frac{1}{\sqrt{x}} + 15x^2 \Rightarrow f'(1) = 1 + 15 = 16\).

c) \(f(x) = (x^2 - 1)(2x^3 + x) \Rightarrow f'(x) = 10x^4 - 3x^2 - 1 \Rightarrow f'(1) = 10 - 3 - 1 = 6\).

d) \(f(x) = \frac{x^2+2}{3x-5} \Rightarrow f'(1) = \frac{3(1)^2 - 10(1) - 6}{9(1)^2 - 30(1) + 25} = \frac{3 - 10 - 6}{9 - 30 + 25} = \frac{-13}{4}\).

e) \(f(x) = (2x^2 - 3x + 8)^{10} \Rightarrow f'(1) = (40 - 30)(2 - 3 + 8)^9 = 10 \cdot 7^9\).

📘 Example 8 : Differentiability at a Point

Determine \(a\) and \(b\) so that \(f(x) = \begin{cases} x^2, & x < 1 \\ ax + b, & x \ge 1 \end{cases}\) is differentiable at \(x = 1\).

Solution: For differentiability at \(x=1\), the function must be continuous and the left/right derivatives must match.

Continuity: \(\lim_{x \to 1^-} f(x) = f(1) \Rightarrow 1^2 = a(1) + b \Rightarrow a + b = 1\).

Derivatives: \(f'(1^-) = 2(1) = 2\), \(f'(1^+) = a\). So \(a = 2\).

Then \(b = 1 - a = -1\). Thus \(a = 2\), \(b = -1\).

📘 Example 9 : Derivative of Square Root Function

Find the derivative of \(y = \sqrt{x^3 + 2x - 1}\).

Solution: Using the rule \(\frac{d}{dx}\sqrt{g(x)} = \frac{g'(x)}{2\sqrt{g(x)}}\), we have:

\[ y' = \frac{3x^2 + 2}{2\sqrt{x^3 + 2x - 1}}. \]

Alternatively, rewrite as \(y = (x^3 + 2x - 1)^{1/2}\) and use the chain rule: \(y' = \frac{1}{2}(x^3 + 2x - 1)^{-1/2} \cdot (3x^2 + 2) = \frac{3x^2 + 2}{2\sqrt{x^3 + 2x - 1}}\).

📘 Example 10 : Derivative of a Square Root of a Power

Find the derivative of \(y = \sqrt{(x^3 + 2x - 1)^3}\).

Solution: Rewrite as \(y = (x^3 + 2x - 1)^{3/2}\). Then using the chain rule:

\[ y' = \frac{3}{2}(x^3 + 2x - 1)^{1/2} \cdot (3x^2 + 2) = \frac{3}{2}(3x^2 + 2)\sqrt{x^3 + 2x - 1}. \]

Proofs of Basic Derivative Rules

Proof of Rule i: \(y = k \Rightarrow y' = 0\)

Let \(f(x) = k\). Then:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{k - k}{h} = \lim_{h \to 0} \frac{0}{h} = 0. \]

Thus, the derivative of any constant function is zero.

Proof of Rule ii: \(y = ax^n \Rightarrow y' = n a x^{n-1}\)

Let \(f(x) = ax^n\). Using the binomial expansion:

\[ (x+h)^n = x^n + n x^{n-1}h + \binom{n}{2} x^{n-2}h^2 + \dots + h^n \]

Then:

\[ \begin{align*} f'(x) &= \lim_{h \to 0} \frac{a(x+h)^n - ax^n}{h} \\ &= \lim_{h \to 0} \frac{a[x^n + n x^{n-1}h + \binom{n}{2} x^{n-2}h^2 + \dots + h^n] - ax^n}{h} \\ &= \lim_{h \to 0} \frac{a n x^{n-1}h + a \binom{n}{2} x^{n-2}h^2 + \dots + a h^n}{h} \\ &= \lim_{h \to 0} \left( a n x^{n-1} + a \binom{n}{2} x^{n-2}h + \dots + a h^{n-1} \right) = a n x^{n-1}. \end{align*} \]

Thus, \(\frac{d}{dx}(ax^n) = n a x^{n-1}\).

Proof of Rule iii: Sum/Difference Rule

Let \(f(x) = U(x) \pm V(x)\). Then:

\[ \begin{align*} f'(x) &= \lim_{h \to 0} \frac{[U(x+h) \pm V(x+h)] - [U(x) \pm V(x)]}{h} \\ &= \lim_{h \to 0} \frac{U(x+h)-U(x)}{h} \pm \lim_{h \to 0} \frac{V(x+h)-V(x)}{h} = U'(x) \pm V'(x). \end{align*} \]

Therefore, the derivative of a sum/difference is the sum/difference of the derivatives.

Proof of Rule iv: Product Rule

Let \(f(x) = U(x) \cdot V(x)\). Then:

\[ \begin{align*} f'(x) &= \lim_{h \to 0} \frac{U(x+h)V(x+h) - U(x)V(x)}{h} \\ &= \lim_{h \to 0} \frac{U(x+h)V(x+h) - U(x+h)V(x) + U(x+h)V(x) - U(x)V(x)}{h} \\ &= \lim_{h \to 0} \left[ U(x+h) \frac{V(x+h)-V(x)}{h} + V(x) \frac{U(x+h)-U(x)}{h} \right] \\ &= U(x) V'(x) + V(x) U'(x). \end{align*} \]

Thus, \(\frac{d}{dx}[U(x)V(x)] = U'(x)V(x) + U(x)V'(x)\).

Proof of Rule v: Quotient Rule

Let \(f(x) = \frac{U(x)}{V(x)}\) with \(V(x) \neq 0\). Then:

\[ \begin{align*} f'(x) &= \lim_{h \to 0} \frac{\frac{U(x+h)}{V(x+h)} - \frac{U(x)}{V(x)}}{h} \\ &= \lim_{h \to 0} \frac{U(x+h)V(x) - U(x)V(x+h)}{h \cdot V(x)V(x+h)} \\ &= \lim_{h \to 0} \frac{[U(x+h)V(x) - U(x)V(x)] + [U(x)V(x) - U(x)V(x+h)]}{h \cdot V(x)V(x+h)} \\ &= \lim_{h \to 0} \frac{V(x)[U(x+h)-U(x)] - U(x)[V(x+h)-V(x)]}{h \cdot V(x)V(x+h)} \\ &= \frac{V(x)U'(x) - U(x)V'(x)}{[V(x)]^2}. \end{align*} \]

Thus, \(\frac{d}{dx}\left[\frac{U(x)}{V(x)}\right] = \frac{U'(x)V(x) - U(x)V'(x)}{[V(x)]^2}\).

Note: For practical purposes, we use these basic rules instead of the limit definition to compute derivatives efficiently. The chain rule (rules vi and vii) is particularly useful for composite functions and will be explored further in the next articles.