We will differentiate $y = \sqrt{\frac{a^2 - x^2}{a^2 + x^2}}$ with respect to $x$.

Step-by-step solution:

Step 1: Rewrite the equation

The given function is: $y = \sqrt{\frac{a^2 - x^2}{a^2 + x^2}}$ Let $u = \frac{a^2 - x^2}{a^2 + x^2}$. Then, $y = \sqrt{u}$.

Step 2: Differentiate $y = \sqrt{u}$

The derivative of $y = \sqrt{u}$ is: $\frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}.$

Step 3: Differentiate $u = \frac{a^2 - x^2}{a^2 + x^2}$

Use the quotient rule: $\frac{du}{dx} = \frac{(a^2 + x^2) \cdot \frac{d}{dx}(a^2 - x^2) - (a^2 - x^2) \cdot \frac{d}{dx}(a^2 + x^2)}{(a^2 + x^2)^2}.$ Simplify each derivative:

• $\frac{d}{dx}(a^2 - x^2) = -2x$,
• $\frac{d}{dx}(a^2 + x^2) = 2x$.

Substitute these into the quotient rule: $\frac{du}{dx} = \frac{(a^2 + x^2)(-2x) - (a^2 - x^2)(2x)}{(a^2 + x^2)^2}.$

Simplify the numerator:

= -2x(a^2 + x^2 + a^2 - x^2) = -2x(2a^2).$$ Thus: $$\frac{du}{dx} = \frac{-4xa^2}{(a^2 + x^2)^2}.$$ #### Step 4: Substitute back into $$\frac{dy}{dx}$$ Now substitute $$\frac{du}{dx}$$ and $$u = \frac{a^2 - x^2}{a^2 + x^2}$$ into $$\frac{dy}{dx}$$: $$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}} \cdot \frac{-4xa^2}{(a^2 + x^2)^2}.$$ Simplify: $$\frac{dy}{dx} = \frac{-4xa^2}{2\sqrt{\frac{a^2 - x^2}{a^2 + x^2}} \cdot (a^2 + x^2)^2}.$$ Combine terms: $$\frac{dy}{dx} = \frac{-2xa^2}{(a^2 + x^2) \sqrt{(a^2 - x^2)(a^2 + x^2)}}.$$ ### Final Answer: $$\frac{dy}{dx} = \frac{-2xa^2}{(a^2 + x^2) \sqrt{a^4 - x^4}}.$$ Would you like me to clarify or expand further? ### Related Questions: 1. How is the chain rule applied in this problem? 2. Can you explain the quotient rule in more detail? 3. What are the limitations of this derivative when $$x \to \pm a$$? 4. How does this derivative behave as $$x \to \infty$$? 5. Could we rewrite the expression in terms of trigonometric functions for a simpler form? **Tip:** Always simplify radicals and fractions separately to avoid confusion in multi-step derivatives.