91Ó°ÊÓ

The product rule is a fundamental concept in calculus. It helps us differentiate functions that are products of two simpler functions. If you have a function described as the product of two sub-functions, such as \( u(x) \) and \( v(x) \), the derivative of this function can be found using the product rule.
The formula for the product rule is:

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

In simple terms, to differentiate \( u(x)v(x) \), differentiate \( u(x) \) while keeping \( v(x) \) unchanged, and vice versa, then add the results together.

For the problem \( xy=12 \), we applied this rule by considering \( x \) and \( y \) analogous to \( u \) and \( v \). Hence, using the product rule, the derivative is \( y + x \frac{dy}{dx} \). This sets the foundation for solving the implicit differentiation problem.

Solving calculus problems often requires a strategic approach. For problems like finding \( \frac{dy}{dx} \) given an implicit equation, the first step is typically to differentiate each term with respect to a variable, often \(x\).
Here's a general approach for these types of problems:

• Identify terms that are products or compositions of functions.
• Use appropriate calculus rules like the product or chain rule.
• Re-arrange the resulting equations to isolate the desired derivative.
• Substitute any given specific values to find the solution.

By systematically applying these steps, you can tackle a variety of challenging problems. In the solution provided earlier, after differentiating using the product rule, rearranging terms allowed us to solve for \( \frac{dy}{dx} \) explicitly.

Differential equations involve functions and their derivatives, offering a way to describe various physical phenomena. An implicit differentiation scenario, such as \( xy = 12 \), is an equation where both variables need consideration to express change.
For simple differential equations arriving from implicit differentiation, remember:

• Both sides of the equation are differentiated.
• The derivative provides the rate of change ratio between the variables; here, that's \( \frac{dy}{dx} = -\frac{y}{x} \).
• This relation can be evaluated for specific values to find exact solutions.

In the exercise above, substituting \( x = 6 \) and \( y = 2 \) into \( -\frac{y}{x} \) allows you to find \( \frac{dy}{dx} \) at that point, solidifying your understanding of their interaction as grounded in the differential equation.