# Nakafa Framework: LLM

URL: /en/subject/high-school/12/mathematics/derivative-function/derivative-of-algebraic-function
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/12/mathematics/derivative-function/derivative-of-algebraic-function/en.mdx

Output docs content for large language models.

---

export const metadata = {
  title: "Derivative of Algebraic Function",
  description: "Master differentiation of algebraic functions using power, product, and quotient rules. Learn to differentiate polynomials, radicals, and rational functions.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/26/2025",
  subject: "Derivative Functions",
};

## Applying the Derivative Rules

Now that we have mastered the various [properties of derivatives](/subject/high-school/12/mathematics/derivative-function/properties-of-derivative-function), it's time to apply them to different forms of algebraic functions. Whether it's a polynomial, a rational function, or one containing a radical, the key is to recognize the function's structure and choose the right 'tool' or property to differentiate it. Let's see how this strategy is applied in a few examples.

## Using the Properties of Derivatives

Let's see how the properties of derivatives work in practice through a few examples.

### Differentiating a Polynomial

Find the first derivative of <InlineMath math="y = 4x^2 + x + 2" />.

**Solution:**

We can differentiate each term one by one using the power rule and the constant rule.

<div className="flex flex-col gap-4">
  <BlockMath math="y' = (4 \cdot 2)x^{2-1} + (1 \cdot 1)x^{1-1} + 0" />
  <BlockMath math="y' = 8x^1 + 1x^0" />
  <BlockMath math="y' = 8x + 1 \quad (\text{since } x^0=1)" />
</div>

### Conquering Radical Forms

Find the first derivative of <InlineMath math="y = x\sqrt{x^3}" />.

**Solution:**

There are two ways to solve this.

**Method 1: Using the Product Rule**

First, we convert the radical form to an exponent: <InlineMath math="y = x \cdot x^{3/2}" />.

Let <InlineMath math="u(x) = x" /> and <InlineMath math="v(x) = x^{3/2}" />. Then, <InlineMath math="u'(x)=1" /> and <InlineMath math="v'(x) = \frac{3}{2}x^{1/2}" />.

<div className="flex flex-col gap-4">
  <BlockMath math="y' = u'(x)v(x) + u(x)v'(x)" />
  <BlockMath math="y' = (1)(x^{3/2}) + (x)(\frac{3}{2}x^{1/2})" />
  <BlockMath math="y' = x^{3/2} + \frac{3}{2}x^{3/2}" />
  <BlockMath math="y' = \frac{5}{2}x^{3/2}" />
</div>

**Method 2: Simplifying First**

We can simplify the function before differentiating it.

<div className="flex flex-col gap-4">
  <BlockMath math="y = x \cdot x^{3/2} = x^{1 + 3/2} = x^{5/2}" />
  <BlockMath math="y' = \frac{5}{2}x^{5/2 - 1} = \frac{5}{2}x^{3/2}" />
</div>

> Both methods give the same result. Sometimes, simplifying the function first can make the differentiation process quicker.

### Tackling Rational Functions

Find the first derivative of <InlineMath math="y = 2\left(\frac{x^2+2}{x}\right)" />.

**Solution:**

We use the quotient rule. Let <InlineMath math="u(x) = x^2+2" /> and <InlineMath math="v(x) = x" />. Then <InlineMath math="u'(x) = 2x" /> and <InlineMath math="v'(x) = 1" />.

<div className="flex flex-col gap-4">
  <BlockMath math="y' = 2 \cdot \left(\frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\right)" />
  <BlockMath math="y' = 2 \cdot \left(\frac{(2x)(x) - (x^2+2)(1)}{x^2}\right)" />
  <BlockMath math="y' = 2 \cdot \left(\frac{2x^2 - x^2 - 2}{x^2}\right)" />
  <BlockMath math="y' = 2 \cdot \left(\frac{x^2 - 2}{x^2}\right)" />
  <BlockMath math="y' = 2 \cdot \left(\frac{x^2}{x^2} - \frac{2}{x^2}\right) = 2 \cdot \left(1 - \frac{2}{x^2}\right) = 2 - \frac{4}{x^2}" />
</div>

## Exercises

1.  Find the first derivative of <InlineMath math="y = \frac{\sqrt{x} + x^2}{2x}" />.

### Answer Key

1.  For this problem, the easiest way is to simplify the function before differentiating it.

    **Step 1: Split the Fraction**
    
    We can break the fraction into two separate parts to make it easier.
    
    <BlockMath math="y = \frac{\sqrt{x}}{2x} + \frac{x^2}{2x}" />

    **Step 2: Simplify Each Term**
    
    Convert the square root to the exponent <InlineMath math="x^{1/2}" /> and use exponent properties to simplify each term.
    
    <div className="flex flex-col gap-4">
      <BlockMath math="y = \frac{x^{1/2}}{2x^1} + \frac{x^2}{2x^1}" />
      <BlockMath math="y = \frac{1}{2}x^{1/2 - 1} + \frac{1}{2}x^{2-1}" />
      <BlockMath math="y = \frac{1}{2}x^{-1/2} + \frac{1}{2}x" />
    </div>

    **Step 3: Apply the Power Rule**
    
    Once the function is simplified, we can directly differentiate it term by term.
    
    <div className="flex flex-col gap-4">
      <BlockMath math="y' = \frac{1}{2}\left(-\frac{1}{2}x^{-1/2 - 1}\right) + \frac{1}{2}(1)" />
      <BlockMath math="y' = -\frac{1}{4}x^{-3/2} + \frac{1}{2}" />
    </div>
    
    So, the first derivative is <InlineMath math="\frac{1}{2} - \frac{1}{4}x^{-3/2}" />.