The equation in the image is:

$x \sqrt{1 - y^2} + y \sqrt{1 - x^2} = C$

We are tasked to prove that:

$\frac{dy}{dx} = - \frac{y}{x} \cdot \frac{\sqrt{1 - x^2}}{\sqrt{1 - y^2}}$

Steps to Prove:

1. Implicit Differentiation:

   • Since $x$ and $y$ are related by the given equation, we use implicit differentiation with respect to $x$.

   The equation is: $x \sqrt{1 - y^2} + y \sqrt{1 - x^2} = C$

2. Differentiate Both Sides: Differentiating both sides with respect to $x$, we get: $\frac{d}{dx} \left( x \sqrt{1 - y^2} \right) + \frac{d}{dx} \left( y \sqrt{1 - x^2} \right) = 0$

   Now, apply the product rule to both terms.

3. Differentiate the First Term $x \sqrt{1 - y^2}$:

   • Product rule: $\frac{d}{dx}(uv) = u'v + uv'$, where $u = x$ and $v = \sqrt{1 - y^2}$. $\frac{d}{dx}(x \sqrt{1 - y^2}) = 1 \cdot \sqrt{1 - y^2} + x \cdot \frac{d}{dx}\left( \sqrt{1 - y^2} \right)$

   The derivative of $\sqrt{1 - y^2}$ with respect to $x$ is: $\frac{d}{dx} \left( \sqrt{1 - y^2} \right) = \frac{-y \frac{dy}{dx}}{\sqrt{1 - y^2}}$

   Therefore, the derivative of the first term is: $\sqrt{1 - y^2} - \frac{xy \frac{dy}{dx}}{\sqrt{1 - y^2}}$

4. Differentiate the Second Term $y \sqrt{1 - x^2}$: Again, applying the product rule, where $u = y$ and $v = \sqrt{1 - x^2}$: $\frac{d}{dx} (y \sqrt{1 - x^2}) = \frac{dy}{dx} \cdot \sqrt{1 - x^2} + y \cdot \frac{-x}{\sqrt{1 - x^2}}$ Simplifying the second term: $= \frac{dy}{dx} \cdot \sqrt{1 - x^2} - \frac{yx}{\sqrt{1 - x^2}}$

5. Combine the Two Results: Now, combine the derivatives of the two terms: $\left( \sqrt{1 - y^2} - \frac{xy \frac{dy}{dx}}{\sqrt{1 - y^2}} \right) + \left( \frac{dy}{dx} \cdot \sqrt{1 - x^2} - \frac{yx}{\sqrt{1 - x^2}} \right) = 0$

6. Solve for $\frac{dy}{dx}$: Collect terms involving $\frac{dy}{dx}$ on one side: $- \frac{xy \frac{dy}{dx}}{\sqrt{1 - y^2}} + \frac{dy}{dx} \cdot \sqrt{1 - x^2} = \frac{yx}{\sqrt{1 - x^2}} - \sqrt{1 - y^2}$

   Factor out $\frac{dy}{dx}$ on the left-hand side: $\frac{dy}{dx} \left( - \frac{xy}{\sqrt{1 - y^2}} + \sqrt{1 - x^2} \right) = \frac{yx}{\sqrt{1 - x^2}} - \sqrt{1 - y^2}$

   This can then be simplified to solve for $\frac{dy}{dx}$, yielding the desired result: $\frac{dy}{dx} = - \frac{y}{x} \cdot \frac{\sqrt{1 - x^2}}{\sqrt{1 - y^2}}$

Would you like further clarification on any part of this derivation?

5 Related Questions:

1. What is implicit differentiation and when is it used?
2. How do we apply the product rule in calculus?
3. Why is the chain rule important in differentiating composite functions?
4. How do we handle differentiation of square roots like $\sqrt{1 - y^2}$?
5. What techniques can we use to simplify implicit derivatives?

Tip:

Always remember to check for opportunities to apply the product and chain rules when differentiating expressions with multiple variables!