2.3 Q-11

Question Statement

Differentiate the following function with respect to $x$:
$y = \frac{2x - 1}{\sqrt{x^2 + 1}}$

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Background and Explanation

This problem involves the differentiation of a rational function containing a square root in the denominator. To solve this, we apply the quotient rule and use chain rule techniques to handle the square root term.

Key Differentiation Rules:

1. Quotient Rule:
   If $y = \frac{u}{v}$, then:
   $\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$
2. Chain Rule for Square Roots:
   If $f(x) = \sqrt{g(x)}$, then:
   $\frac{d}{dx}\sqrt{g(x)} = \frac{1}{2\sqrt{g(x)}} \cdot \frac{dg(x)}{dx}$
3. Simplifying algebraic expressions is critical for clarity in the solution.

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Solution

Step 1: Identify the numerator and denominator

The function is:
$y = \frac{2x - 1}{\sqrt{x^2 + 1}}$
Here:

• Numerator: $u = 2x - 1$
• Denominator: $v = \sqrt{x^2 + 1}$

Step 2: Apply the quotient rule

Using the quotient rule:
$\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$

a) Differentiate $u = 2x - 1$:

$\frac{du}{dx} = 2$

b) Differentiate $v = \sqrt{x^2 + 1}$:

Using the chain rule: $\frac{dv}{dx} = \frac{1}{2\sqrt{x^2 + 1}} \cdot \frac{d}{dx}(x^2 + 1)$ $\frac{dv}{dx} = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x$ $\frac{dv}{dx} = \frac{x}{\sqrt{x^2 + 1}}$

Step 3: Substitute into the quotient rule

Substitute $u$, $v$, and their derivatives: $\frac{dy}{dx} = \frac{\sqrt{x^2 + 1} \cdot 2 - (2x - 1) \cdot \frac{x}{\sqrt{x^2 + 1}}}{(\sqrt{x^2 + 1})^2}$

Step 4: Simplify the numerator

Expand the terms in the numerator:

1. First term: $2\sqrt{x^2 + 1}$
2. Second term: $(2x - 1) \cdot \frac{x}{\sqrt{x^2 + 1}} = \frac{x(2x - 1)}{\sqrt{x^2 + 1}}$

Combine terms: $\text{Numerator} = 2\sqrt{x^2 + 1} - \frac{x(2x - 1)}{\sqrt{x^2 + 1}}$
Express the numerator over a common denominator: $\text{Numerator} = \frac{2(x^2 + 1) - x(2x - 1)}{\sqrt{x^2 + 1}}$

Expand and simplify: $\text{Numerator} = \frac{2x^2 + 2 - 2x^2 + x}{\sqrt{x^2 + 1}}$
$\text{Numerator} = \frac{x + 2}{\sqrt{x^2 + 1}}$

Step 5: Write the final expression

The denominator is $(\sqrt{x^2 + 1})^2 = x^2 + 1$. The derivative becomes: $\frac{dy}{dx} = \frac{x + 2}{(x^2 + 1)^{3/2}}$

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Key Formulas or Methods Used

1. Quotient Rule:
   $\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$
2. Chain Rule for Square Roots:
   $\frac{d}{dx}\sqrt{g(x)} = \frac{1}{2\sqrt{g(x)}} \cdot \frac{dg(x)}{dx}$
3. Simplification of algebraic fractions.

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Summary of Steps

1. Identify $u = 2x - 1$ and $v = \sqrt{x^2 + 1}$.
2. Differentiate $u$ and $v$:
   • $\frac{du}{dx} = 2$
   • $\frac{dv}{dx} = \frac{x}{\sqrt{x^2 + 1}}$
3. Apply the quotient rule: $\frac{dy}{dx} = \frac{\sqrt{x^2 + 1} \cdot 2 - (2x - 1) \cdot \frac{x}{\sqrt{x^2 + 1}}}{(\sqrt{x^2 + 1})^2}$
4. Simplify the numerator: $\text{Numerator} = \frac{x + 2}{\sqrt{x^2 + 1}}$
5. Write the final expression: $\frac{dy}{dx} = \frac{x + 2}{(x^2 + 1)^{3/2}}$