2.3 Q-14

Question Statement

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

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Background and Explanation

This problem involves differentiating a fraction of two square roots. To solve it, we will use the following:

1. Algebraic Simplification: First simplify the expression by multiplying the numerator and denominator by a conjugate to eliminate square roots.
2. Quotient Rule: Once simplified, apply the quotient rule to differentiate the resulting expression.

Key Rules:

1. Quotient Rule:
   If $f(x) = \frac{u}{v}$, then:
   $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$

2. Chain Rule for Square Roots:
   If $f(x) = \sqrt{g(x)}$, then:
   $\frac{d}{dx} \sqrt{g(x)} = \frac{1}{2 \sqrt{g(x)}} \cdot \frac{dg(x)}{dx}$

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Solution

Step 1: Simplify the expression using the conjugate

Start with the given expression: $y = \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$

Multiply both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{1+x} - \sqrt{1-x}$: $y = \frac{(\sqrt{1+x} - \sqrt{1-x})^2}{(\sqrt{1+x})^2 - (\sqrt{1-x})^2}$

Step 2: Simplify the numerator and denominator

• Numerator: Expand $(\sqrt{1+x} - \sqrt{1-x})^2$: $(\sqrt{1+x} - \sqrt{1-x})^2 = (1+x) + (1-x) - 2 \sqrt{(1+x)(1-x)}$
  Simplifying: $= 2 - 2 \sqrt{1 - x^2}$

• Denominator: Simplify using the difference of squares formula: $(\sqrt{1+x})^2 - (\sqrt{1-x})^2 = (1+x) - (1-x) = 2x$

So, we have: $y = \frac{2 - 2 \sqrt{1 - x^2}}{2x}$

Step 3: Simplify the expression further

Factor out the common factor of 2 in the numerator: $y = \frac{2(1 - \sqrt{1 - x^2})}{2x}$

Cancel the 2s: $y = \frac{1 - \sqrt{1 - x^2}}{x}$

Now, we have a simpler expression for $y$: $y = \frac{1 - \sqrt{1 - x^2}}{x}$

Step 4: Differentiate the simplified expression

Now, differentiate $y = \frac{1 - \sqrt{1 - x^2}}{x}$ with respect to $x$ using the quotient rule.

Let:

• $u = 1 - \sqrt{1 - x^2}$
• $v = x$

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

Step 4.1: Find $\frac{du}{dx}$ and $\frac{dv}{dx}$

• For $u = 1 - \sqrt{1 - x^2}$, use the chain rule: $\frac{du}{dx} = -\frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{x}{\sqrt{1 - x^2}}$

• For $v = x$, we have: $\frac{dv}{dx} = 1$

Step 4.2: Apply the quotient rule

Substitute into the quotient rule formula: $\frac{dy}{dx} = \frac{x \cdot \frac{x}{\sqrt{1 - x^2}} - (1 - \sqrt{1 - x^2}) \cdot 1}{x^2}$

Simplify the numerator: $= \frac{x^2/\sqrt{1 - x^2} - (1 - \sqrt{1 - x^2})}{x^2}$

Step 5: Simplify further

Simplify the expression: $= \frac{x^2 - \sqrt{1 - x^2} + 1 - x^2}{x^2 \sqrt{1 - x^2}}$

Now cancel the $x^2$ terms: $= \frac{1 - \sqrt{1 - x^2}}{x^2 \sqrt{1 - x^2}}$

Thus, the final derivative is: $\frac{dy}{dx} = \frac{1 - \sqrt{1 - x^2}}{x^2 \sqrt{1 - x^2}}$

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Key Formulas or Methods Used

1. Quotient Rule:
   $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$

2. Chain Rule for Square Roots:
   $\frac{d}{dx} \sqrt{g(x)} = \frac{1}{2 \sqrt{g(x)}} \cdot \frac{dg(x)}{dx}$

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Summary of Steps

1. Simplify the expression by multiplying the numerator and denominator by the conjugate.
2. Expand and simplify the numerator and denominator.
3. Simplify the resulting expression.
4. Use the quotient rule to differentiate the simplified expression.
5. Final derivative: $\frac{dy}{dx} = \frac{1 - \sqrt{1 - x^2}}{x^2 \sqrt{1 - x^2}}$