2.3 Q-15

Question Statement

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

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Background and Explanation

This problem involves differentiating a fraction with square roots. The key concepts to be used here include:

1. Product Rule: Since the expression is a product of two functions ($x$ and $\frac{\sqrt{a+x}}{\sqrt{a-x}}$), we will apply the product rule.
2. Quotient Rule: The second term involves differentiating a quotient of two functions, which requires the quotient rule.
3. Chain Rule: When differentiating the square roots in the expression, we will need the chain rule.

Key Rules:

1. Product Rule:
   If $f(x) = u(x) \cdot v(x)$, then:
   $\frac{d}{dx} \left( u(x) \cdot v(x) \right) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

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

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

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Solution

Step 1: Apply the Product Rule

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

The expression is the product of two functions: $x$ and $\frac{\sqrt{a+x}}{\sqrt{a-x}}$. We apply the product rule: $\frac{d y}{d x} = \frac{d}{d x} \left( x \right) \cdot \frac{\sqrt{a+x}}{\sqrt{a-x}} + x \cdot \frac{d}{d x} \left( \frac{\sqrt{a+x}}{\sqrt{a-x}} \right)$

• The derivative of $x$ is 1, so the first term simplifies to: $\frac{\sqrt{a+x}}{\sqrt{a-x}}$

Step 2: Differentiate the Quotient $\frac{\sqrt{a+x}}{\sqrt{a-x}}$

Now we differentiate the second term: $x \cdot \frac{d}{d x} \left( \frac{\sqrt{a+x}}{\sqrt{a-x}} \right)$

We apply the quotient rule for the derivative of a fraction: $\frac{d}{d x} \left( \frac{\sqrt{a+x}}{\sqrt{a-x}} \right) = \frac{\sqrt{a-x} \cdot \frac{d}{dx} (\sqrt{a+x}) - \sqrt{a+x} \cdot \frac{d}{dx} (\sqrt{a-x})}{(\sqrt{a-x})^2}$

Step 2.1: Apply the Chain Rule to the Square Roots

• For $\frac{d}{dx} \sqrt{a+x}$: $\frac{d}{dx} \sqrt{a+x} = \frac{1}{2\sqrt{a+x}}$

• For $\frac{d}{dx} \sqrt{a-x}$: $\frac{d}{dx} \sqrt{a-x} = \frac{-1}{2\sqrt{a-x}}$

Step 2.2: Substitute These Derivatives

Now, substitute the derivatives back into the quotient rule formula: $\frac{d}{d x} \left( \frac{\sqrt{a+x}}{\sqrt{a-x}} \right) = \frac{\sqrt{a-x} \cdot \frac{1}{2\sqrt{a+x}} - \sqrt{a+x} \cdot \frac{-1}{2\sqrt{a-x}}}{a-x}$

Simplifying: $= \frac{\frac{a-x}{2\sqrt{a+x}\sqrt{a-x}} + \frac{\sqrt{a+x}}{2\sqrt{a-x}}}{a-x}$

Step 3: Simplify the Result

Now, simplify the expression further. We can cancel terms and combine fractions: $= \frac{(a-x)(a+x)}{2 \sqrt{a+x} \cdot \sqrt{a-x} \cdot (a-x)}$

Finally, simplifying the expression results in: $= \frac{a}{2 \sqrt{a+x} \cdot \sqrt{a-x} \cdot (a-x)}$

Step 4: Combine All Terms

Now, combine all the terms: $\frac{d y}{d x} = \frac{\sqrt{a+x}}{\sqrt{a-x}} + x \cdot \frac{a}{2 \sqrt{a+x} \cdot \sqrt{a-x} \cdot (a-x)}$

This can be rewritten as: $\frac{d y}{d x} = \frac{(a+x)(a-x) + a x}{(a-x)^{3/2} \cdot \sqrt{a+x}}$

Step 5: Final Simplified Expression

Finally, simplifying the numerator gives: $= \frac{a^2 - x^2 + a x}{(a+x)^{1/2} (a-x)^{3/2}}$

Thus, the final derivative is: $\frac{dy}{dx} = \frac{a^2 - x^2 + a x}{(a+x)^{1/2} (a-x)^{3/2}}$

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Key Formulas or Methods Used

1. Product Rule:
   $\frac{d}{dx} \left( u(x) \cdot v(x) \right) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

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

3. 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. Apply the product rule to the given expression.
2. Differentiate the quotient using the quotient rule.
3. Apply the chain rule to differentiate the square roots.
4. Simplify the result from the quotient rule.
5. Combine all terms and simplify the final expression.
6. The final derivative is: $\frac{dy}{dx} = \frac{a^2 - x^2 + a x}{(a+x)^{1/2} (a-x)^{3/2}}$