91Ó°ÊÓ

Calculus is a branch of mathematics that focuses on change and motion. It is divided into two main parts: differential calculus and integral calculus. In this context, we are focused on differential calculus.

• Differential Calculus: This involves studying rates at which quantities change. It helps to find slopes of curves or the derivative of a function.
• Integral Calculus: This deals with accumulation of quantities, such as areas under a curve.

Implicit differentiation, as used in this problem, is a technique in differential calculus. It allows us to find derivatives of functions that are not isolated, or explicitly solved, for one variable. Often, in real-world applications, variables are intertwined, necessitating implicit differentiation. By differentiating both sides of an equation with respect to a chosen variable, we can unveil relationships between the changing rates of the involved variables.

Derivatives represent the rate of change of a function relative to a variable. If you're standing on a hill, the derivative would tell you how steep the hill is at that particular point.

• Notation: Derivatives are often denoted by \(f'(x)\), \(\frac{d}{dx}f(x)\), or for implicitly related variables, \(\frac{dy}{dx}\).
• Calculation: To calculate derivatives, rules like the power rule, product rule, and chain rule are used extensively.

In this exercise, we see the use of differentiation for both explicit and implicit functions. On one hand, differentiating \(x^4\) is explicit and straightforward: it yields \(4x^3\). But on the other hand, \(\sin y\) involves \(y\), requiring implicit differentiation. Thus, its derivative becomes \(\cos y \cdot \frac{dy}{dx}\), indicating the relationship between \(x\) and \(y\) is not independent.

The product rule is a crucial derivative tool when dealing with the product of two functions. It is especially useful when functions are multiplied together, as they often are in real-world problems. For two functions \((u)\) and \((v)\), the product rule is expressed as:\[\frac{d}{dx}(uv) = u'v + uv'\]In the example provided, the right side of the equation \(x^3 y^2\) needs the product rule. This is because it is composed of two functions, \((x^3)\) and \((y^2)\). Applying the product rule, we find:- Derivative of \((x^3)\) is \(3x^2\).- Derivative of \((y^2)\) is \(2y \, \frac{dy}{dx}\).Consequently, differentiating \(x^3 y^2\) yields \(3x^2 y^2 + x^3 \cdot 2y \, \frac{dy}{dx}\). This plays a vital role in arriving at the final implicit derivative, as it allows us to separate and isolate terms involving \(\frac{dy}{dx}\). Understanding and applying the product rule is crucial, as it helps tackle complex differentiation problems effectively.