To find the derivative of $y$ with respect to $x$ from the given equation

$(x^2 - y^2)^3 = 3a^4x^2,$

we'll use implicit differentiation. Here are the steps:

Step 1: Differentiate both sides with respect to $x$

We'll apply the chain rule on the left-hand side and differentiate each term on the right-hand side.

For the left-hand side: $\frac{d}{dx} \left( (x^2 - y^2)^3 \right)$ Using the chain rule, we differentiate the outer function (power of 3) and then the inner function:

$3(x^2 - y^2)^2 \cdot \frac{d}{dx} (x^2 - y^2)$

Now, differentiate $(x^2 - y^2)$ using the chain rule:

$\frac{d}{dx}(x^2 - y^2) = 2x - 2y \cdot \frac{dy}{dx}$

So, the derivative of the left-hand side becomes:

$3(x^2 - y^2)^2 \cdot (2x - 2y \cdot \frac{dy}{dx})$

For the right-hand side:

$\frac{d}{dx} (3a^4 x^2) = 6a^4 x$

Step 2: Combine the results

The equation now becomes:

$3(x^2 - y^2)^2 \cdot (2x - 2y \cdot \frac{dy}{dx}) = 6a^4 x$

Step 3: Solve for $\frac{dy}{dx}$

Expand both sides:

$6(x^2 - y^2)^2 \cdot x - 6(x^2 - y^2)^2 \cdot y \cdot \frac{dy}{dx} = 6a^4 x$

Now, collect all terms involving $\frac{dy}{dx}$ on one side:

$-6(x^2 - y^2)^2 \cdot y \cdot \frac{dy}{dx} = 6a^4 x - 6(x^2 - y^2)^2 \cdot x$

Factor out the constants:

$-6y(x^2 - y^2)^2 \cdot \frac{dy}{dx} = 6x \left( a^4 - (x^2 - y^2)^2 \right)$

Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-6x \left( a^4 - (x^2 - y^2)^2 \right)}{6y(x^2 - y^2)^2}$

Simplify:

$\frac{dy}{dx} = \frac{-x \left( a^4 - (x^2 - y^2)^2 \right)}{y(x^2 - y^2)^2}$

Final Answer:

$\frac{dy}{dx} = \frac{-x \left( a^4 - (x^2 - y^2)^2 \right)}{y(x^2 - y^2)^2}$

Would you like further details on any of the steps or have additional questions about implicit differentiation?

Related Questions:

1. What is the chain rule in differentiation?
2. Can you differentiate an implicit function without solving for $y$ explicitly?
3. How do we apply implicit differentiation to trigonometric functions?
4. What is the product rule and how can it be used in implicit differentiation?
5. How do you interpret the derivative $\frac{dy}{dx}$ geometrically?

Tip:

Always remember to treat $y$ as a function of $x$ when differentiating implicitly. The derivative of $y$ with respect to $x$ will include $\frac{dy}{dx}$.