91Ó°ÊÓ

The chain rule is a fundamental tool in calculus, especially when dealing with composite functions. When you have a function within another function, the chain rule helps find the derivative.
The rule states: if you have a function \( f(g(x)) \), then the derivative is \( f'(g(x)) \cdot g'(x) \). This is essential for differentiating functions like \( \sin(x^2y^2) \) in our exercise.
To apply the chain rule here:

• Recognize the inner function \( u = x^2y^2 \)
• Differentiate the outer function, which is \( \sin(u) \), resulting in \( \cos(u) \)
• Multiply by the derivative of the inner function \( u \), using further differentiation

This layered approach helps in meticulously breaking down the derivative process, ensuring that every part of the composite function is accounted for.

The product rule comes into play when differentiating products of two functions. For a product \( f(x) \cdot g(x) \), the product rule states that the derivative is \( f'(x)g(x) + f(x)g'(x) \). This is twin to the chain rule when differentiating something like \( x^2y^2 \).
In our exercise:

• We see the product of \( x^2 \) and \( y^2 \)
• Applying the product rule, differentiate \( x^2 \) to get \( 2x \) and multiply by \( y^2 \)
• Then, differentiate \( y^2 \) using the chain rule, where you treat \( y \) as a function of \( x \)

Together, these steps yield \( 2xy^2 + x^2 \cdot 2y \frac{dy}{dx} \). This part of the solution is crucial to tackling the whole implicit differentiation problem.

Implicit differentiation is a technique used when it's difficult or impossible to solve for \( y \) explicitly. In our problem, \( y \) is tied to \( x \) in a complex way due to it being inside a trigonometric function.
When faced with implicit functions:

• Differentiate both sides of the equation with respect to \( x \)
• Apply relevant differentiation rules such as the chain rule and the product rule
• After differentiating, rearrange the equation to isolate \( \frac{dy}{dx} \)

Implicit differentiation broadens the realm of problems you can solve and is especially helpful for those involving relations rather than functions. Our exercise is a classic example, showing how implicit differentiation simplifies complex relationships, giving us the result \( \frac{dy}{dx} = \frac{1 - 2xy^2 \cos(x^2y^2)}{2x^2y \cos(x^2y^2)} \). By the end, you transform a seemingly daunting equation into a solvable expression for the derivative.