91Ó°ÊÓ

When differentiating expressions involving products of two functions, the product rule is essential. The product rule states: if you have two differentiable functions, say \( u(x) \) and \( v(x) \), their derivative is given by \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \). In simpler terms, differentiate one function while keeping the other constant, then reverse the roles and add the results.

In our exercise, we encountered the term \( -xy \). We treat it as a product of \( x \) and \( y \). Applying the product rule, the derivative becomes:

• Differentiate \( x \) to get 1, and multiply by \( y \): \( 1 \cdot y = y \).
• Keep \( x \) constant and differentiate \( y \), which means you multiply \( x \) by \( \frac{dy}{dx} \): \( x \cdot \frac{dy}{dx} \).

Combining these gives \( -(x \frac{dy}{dx} + y) \), completing the application of the product rule.

The chain rule is a powerful technique used when differentiating composite functions. Simply put, if you have a function nested within another function, you apply the chain rule to find the derivative. Mathematically, if \( z = f(g(x)) \), then the derivative \( \frac{dz}{dx} \) is \( f'(g(x)) \cdot g'(x) \).

In the given problem, the chain rule is required for the term \( y^3 \), where \( y \) itself is a function of \( x \). To differentiate \( y^3 \) with respect to \( x \), we first differentiate \( y^3 \) with respect to \( y \), giving \( 3y^2 \), and then multiply by the derivative of \( y \) with respect to \( x \), which is \( \frac{dy}{dx} \).

This results in the term \( 3y^2 \frac{dy}{dx} \) in our differentiated equation, demonstrating the chain rule in action.

Derivatives are central to calculus and express how a function changes as its input changes. When calculating derivatives of various functions, different rules apply depending on the composition and operations in the function.

Common rules include:

• The power rule: For a term like \( x^n \), the derivative \( nx^{n-1} \).
• The product rule: For the product of two functions, as discussed earlier.
• The chain rule: For nested functions, as detailed previously.

Given our exercise equation \( x^3 - xy + y^3 = 1 \), we applied each of these rules accordingly.

Each part of the original equation was differentiated respecting the specific rules, resulting in an implicit derivative expression: \( 3x^2 - (x \frac{dy}{dx} + y) + 3y^2 \frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \), we obtained \( \frac{dy}{dx} = \frac{y - 3x^2}{3y^2 - x} \), representing the derivative of \( y \) with respect to \( x \).