91Ó°ÊÓ

The product rule is a fundamental tool in calculus when you need to differentiate expressions involving products of two functions. It allows us to tackle complex problems like differentiating the term \( x^3 y^2 \) in our original exercise.For two functions, \( u(x) \) and \( v(x) \), the product rule states:

• \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) \)

In our problem, consider \( u = x^3 \) and \( v = y^2 \). Applying the product rule:- Differentiate \( u(x) = x^3 \) to get \( u'(x) = 3x^2 \).- Differentiate \( v(x) = y^2 \) with respect to \( x \) using the chain rule, giving \( 2y \cdot \frac{dy}{dx} \).Substituting, we have:

• \( u'(x)v(x) = 3x^2 \cdot y^2 \)
• \( u(x)v'(x) = x^3 \cdot 2y \cdot \frac{dy}{dx} \)

Combine these results to get the derivative of \( x^3y^2 \) as part of implicit differentiation for our equation.

The chain rule is an essential calculus principle used to differentiate composite functions, such as in our given problem when differentiating terms that include both \( x \) and \( y \).The chain rule states:

• If a function \( y = f(g(x)) \), then its derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

This rule is especially useful in implicit differentiation where one variable is expressed in terms of another implicitly.In our example, consider differentiating \( y^2 \) with respect to \( x \):- Here, \( y \) is a function of \( x \), so differentiate \( y^2 \) by applying the chain rule: \( 2y \cdot \frac{dy}{dx} \).Such use of the chain rule allows us to effectively handle terms where \( y \) is not explicitly a function of \( x \), a common occurrence in implicit differentiation problems.

Solving calculus problems often requires implementing strategies step-by-step, which is vital for solving the implicit differentiation problem here.Begin by differentiating each term:- Differentiate with respect to \( x \), applying both the product and chain rules where necessary.- For example, differentiate \( x^3y^2 \) and \(-5x^2y\) using the product and chain rules.Then, follow these steps:

• Collect all terms containing \( \frac{dy}{dx} \) on one side to isolate it.
• Factor \( \frac{dy}{dx} \) out from these terms.
• Solve for \( \frac{dy}{dx} \) to find the derivative, using algebra to rearrange the equation if needed.

Precision and attention to detail in each step are crucial. When terms involve both \( x \) and \( y \), accurate application of rules simplify the process, leading to the final solution for \( \frac{dy}{dx} \) as seen in our example problem.