To find the derivative of $y = \sqrt{\frac{a^2 - x^2}{a^2 + x^2}}$, we will use the chain rule and quotient rule systematically.

Step-by-Step Solution:

1. Rewrite the function: $y = \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{1/2}$

2. Apply the chain rule: Let $u = \frac{a^2 - x^2}{a^2 + x^2}$, so $y = u^{1/2}$. Then: $\frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx}$ Substituting $u$, this becomes: $\frac{dy}{dx} = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{-1/2} \cdot \frac{d}{dx}\left( \frac{a^2 - x^2}{a^2 + x^2} \right)$

3. Differentiate $u = \frac{a^2 - x^2}{a^2 + x^2}$ using the quotient rule: Let $u = \frac{v}{w}$, where:

   • $v = a^2 - x^2$, so $\frac{dv}{dx} = -2x$,
   • $w = a^2 + x^2$, so $\frac{dw}{dx} = 2x$. The quotient rule gives: $\frac{du}{dx} = \frac{w \cdot \frac{dv}{dx} - v \cdot \frac{dw}{dx}}{w^2}$ Substituting $v$ and $w$: $\frac{du}{dx} = \frac{(a^2 + x^2)(-2x) - (a^2 - x^2)(2x)}{(a^2 + x^2)^2}$ Simplify the numerator: $(a^2 + x^2)(-2x) - (a^2 - x^2)(2x) = -2x(a^2 + x^2) - 2x(a^2 - x^2)$ $= -2x(a^2 + x^2 + a^2 - x^2) = -2x(2a^2) = -4xa^2$ Thus: $\frac{du}{dx} = \frac{-4xa^2}{(a^2 + x^2)^2}$

4. Substitute back into the derivative: $\frac{dy}{dx} = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{-1/2} \cdot \frac{-4xa^2}{(a^2 + x^2)^2}$

5. Simplify the expression: Rewrite $\left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{-1/2}$ as $\sqrt{\frac{a^2 + x^2}{a^2 - x^2}}$: $\frac{dy}{dx} = \frac{1}{2} \cdot \sqrt{\frac{a^2 + x^2}{a^2 - x^2}} \cdot \frac{-4xa^2}{(a^2 + x^2)^2}$ Combine terms: $\frac{dy}{dx} = \frac{-2xa^2 \cdot \sqrt{a^2 + x^2}}{\sqrt{a^2 - x^2} \cdot (a^2 + x^2)^2}$

Final Answer:

$\frac{dy}{dx} = \frac{-2xa^2 \sqrt{a^2 + x^2}}{\sqrt{a^2 - x^2} \cdot (a^2 + x^2)^2}$

Let me know if you have any questions or want further explanation!

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Questions to deepen understanding:

1. What role does the chain rule play in finding derivatives of composite functions?
2. How is the quotient rule applied in the context of this problem?
3. Why do we rewrite the square root in terms of a fractional exponent?
4. Can the result be simplified further in terms of physical or geometric interpretations?
5. How does the derivative behave as $x$ approaches $\pm a$?

Tip: Always keep track of each step in composite functions to ensure clarity when applying