91Ó°ÊÓ

In calculus, the product rule is essential when you need to differentiate products of two functions. Here, the original equation utilizes the product rule to differentiate the left side of the equation, which is a product of two functions: \( 4 \cos x \) and \( \sin y \). When differentiating a product, the product rule suggests that the derivative of \( u(x)\cdot v(x) \) is \( u'(x)v(x) + u(x)v'(x) \). This means we take the derivative of the first function, multiply it by the second function, and then add the first function multiplied by the derivative of the second function.
Applying this to our problem:

• extbf{First function:} \( 4 \cos x\) - differentiated as \( 4(-\sin x)\)
• extbf{Second function:} \( \sin y \) - keep it as \( \sin y \) initially and then differentiate it using the chain rule.

With these derivatives, you should be able to apply the product rule effectively in implicit differentiation problems by recognizing the terms of a product and their derivatives.

The chain rule is a powerful technique used when differentiating composite functions. It comes into play in our problem during the differentiation of the term \( \sin y \), as it depends on another function \( y \), which is itself a function of \( x \). To differentiate such expressions, the chain rule states that the derivative is found by differentiating the outside function and then multiplying by the derivative of the inner function.
In our context, the outside function is \( \sin y \), and the differentiation gives \( \cos y \). The inner function \( y \) differentiates to \( \frac{dy}{dx} \). Thus, \(\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}\).

• The outside function is differentiated: \( \sin y \) results in \( \cos y \).
• The inside function \( y \) is differentiated concerning \( x \) to give \( \frac{dy}{dx} \).

Understanding the chain rule helps you tackle more complex equations involving nested functions and remains a critical tool in calculus, particularly in implicit differentiation.

Trigonometric identities simplify expressions involving trigonometric functions. In this problem, the last step capitalizes on these identities, specifically the tangent identity, to simplify and express the derivative in a tidy form.

• Recall the tangent identity: \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).

Our equation results in the expression \( \frac{\sin x \sin y}{\cos x \cos y} \). By recognizing \( \tan x = \frac{\sin x}{\cos x} \) and \( \tan y = \frac{\sin y}{\cos y} \), you can rewrite the expression as \(\tan x \tan y \), which is much simpler to understand and use.
Using trigonometric identities like these not only makes the expressions cleaner but also reveals insights into relationships between different trigonometric functions, making problem-solving more intuitive.