2.3 Q-10

Question Statement

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

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Background and Explanation

This is a quotient of two functions, so we will apply the quotient rule. The quotient rule states:
$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$

Here, $u(x)$ and $v(x)$ represent the numerator and denominator functions, respectively. We will also use the derivative of the square root function:
$\frac{d}{dx} \sqrt{f(x)} = \frac{f'(x)}{2\sqrt{f(x)}}$

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Solution

Let: $Y = \frac{\sqrt{1+x}}{\sqrt{1-x}}$

To differentiate $Y$ with respect to $x$, we use the quotient rule: $\frac{dY}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{v(x)^2}$

Here:

• $u(x) = \sqrt{1+x}$ and $v(x) = \sqrt{1-x}$

Step 1: Differentiate $u(x)$

$u(x) = \sqrt{1+x} \implies u'(x) = \frac{1}{2\sqrt{1+x}}$

Step 2: Differentiate $v(x)$

$v(x) = \sqrt{1-x} \implies v'(x) = \frac{-1}{2\sqrt{1-x}}$

Step 3: Apply the quotient rule

Substitute $u(x)$, $u'(x)$, $v(x)$, and $v'(x)$ into the quotient rule: $\frac{dY}{dx} = \frac{\sqrt{1-x} \cdot \frac{1}{2\sqrt{1+x}} - \sqrt{1+x} \cdot \frac{-1}{2\sqrt{1-x}}}{(1-x)}$

Step 4: Simplify the numerator

The numerator becomes: $\sqrt{1-x} \cdot \frac{1}{2\sqrt{1+x}} + \sqrt{1+x} \cdot \frac{1}{2\sqrt{1-x}}$

Combine under a common denominator: $= \frac{(1-x) + (1+x)}{2\sqrt{(1+x)(1-x)}}$

Simplify further: $= \frac{2}{2\sqrt{(1+x)(1-x)}}$

Step 5: Simplify the fraction

The result is: $\frac{dY}{dx} = \frac{1}{\sqrt{1+x} \cdot \sqrt{1-x}}$

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Key Formulas or Methods Used

1. Quotient Rule:
   $\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$
2. Derivative of a Square Root:
   $\frac{d}{dx} \sqrt{f(x)} = \frac{f'(x)}{2\sqrt{f(x)}}$

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Summary of Steps

1. Identify $u(x)$ and $v(x)$:
   • $u(x) = \sqrt{1+x}$ and $v(x) = \sqrt{1-x}$.
2. Differentiate $u(x)$ and $v(x)$:
   • $u'(x) = \frac{1}{2\sqrt{1+x}}$
   • $v'(x) = \frac{-1}{2\sqrt{1-x}}$.
3. Apply the quotient rule to find $\frac{dY}{dx}$.
4. Simplify the expression: $\frac{dy}{dx} = \frac{1}{\sqrt{1+x}(1-x)^{1/2}}$

Final Answer: $\frac{dY}{dx} = \frac{1}{\sqrt{1+x} \cdot \sqrt{1-x}}$