91Ó°ÊÓ

The chain rule is an essential tool in calculus, especially when dealing with composite functions. In the context of implicit differentiation, it helps us differentiate functions of multiple variables, often where one variable is expressed implicitly.
When we apply the chain rule, we are essentially saying: if you have a composition of functions, like \( f(g(x)) \), you can find its derivative by multiplying the derivative of the outer function by the derivative of the inner function. In other words, the chain rule states:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x) \)

In the equation \( x^4 + \sin y = x^3 y^2 \), to differentiate \( \sin y \) with respect to \( x \), we need to use the chain rule. Here, \( y \) is a function of \( x \), so its derivative is \( \cos y \cdot \frac{dy}{dx} \). Using the chain rule allows us to connect these dots and account for the implicit nature of \( y \). This technique is crucial for accurately differentiating terms where \( y \) is intertwined with \( x \) in an indirect way.

The product rule comes into play when differentiating the product of two functions. If you have functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) follows these steps:

• First, differentiate \( u(x) \) while keeping \( v(x) \) the same.
• Second, differentiate \( v(x) \) while keeping \( u(x) \) the same.
• Finally, add these two results together.

In formula terms, this is expressed as:

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

In our problem, we apply the product rule to the expression \( x^3 y^2 \). Treat \( x^3 \) as one function and \( y^2 \) as another. Differentiating \( x^3 \) gives \( 3x^2 \), and \( y^2 \) requires the use of the chain rule since it's a function of \( y(x) \). Hence, the derivative becomes \( x^3 \cdot 2y \cdot \frac{dy}{dx} \). This approach gives you the complete derivative of the product as \( 3x^2 y^2 + x^3 \cdot 2y \cdot \frac{dy}{dx} \), capturing all the interdependencies between \( x \) and \( y \).

Derivatives of trigonometric functions are fundamental to many calculus problems. These functions include sine, cosine, and others that often appear in implicit differentiation issues. Knowing these derivatives is crucial.
The basic derivative rules for the most common trigonometric functions are as follows:

• \( \frac{d}{dx} [\sin x] = \cos x \)
• \( \frac{d}{dx} [\cos x] = -\sin x \)
• \( \frac{d}{dx} [\tan x] = \sec^2 x \)

In the given exercise, we specifically use the derivative of \( \sin y \), which is \( \cos y \). But since \( y \) is a function of \( x \), we must also multiply by \( \frac{dy}{dx} \) due to the implicit differentiation and chain rule.
This encounter with trigonometric functions underscores why having a solid grasp of their derivatives is pivotal, aiding in seamless transitions across different calculus problems.