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The chain rule is a fundamental principle in calculus used to find the derivative of a composite function. To understand the chain rule, imagine you're peeling an onion layer by layer; each function in the composition represents a layer, and as you peel them away, you apply the chain rule successively.

The general formula for the chain rule when dealing with single variable functions is: $$\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$$Now, when it comes to partial derivatives, the chain rule is slightly adapted. For instance, with the function \( z = f(x, y) \), where \( x \) and \( y \) are functions of another variable (like \( t \)), the chain rule tells us how to find the derivative of \( z \) with respect to that variable:
$$\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$$
This approach is applied when computing the first partial derivative of \( z = \sin 3x \cos 3y \) with respect to \( x \). Here we treat \( y \) as a constant, apply the chain rule and subsequently use the knowledge that the derivative of \( \sin \) is \( \cos \), giving us \( 3 \cos(3x) \cos(3y) \).

The product rule is another essential technique used to differentiate functions that are multiplied by one another. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), their derivative is not simply the product of their individual derivatives. Instead, the product rule formula goes as follows:
$$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$
This formula ensures that we take into account how each function changes in relation to the other. This mathematical concept is crucial when we look at the original exercise problem, where we deal with \( z = \sin 3x \cos 3y \). To find \( \frac{\partial z}{\partial y} \), we again view \( x \) as constant and apply the product rule to the remaining functions of \( y \). The differentiation of \( \sin(3x) \cos(3y) \) with respect to \( y \) requires us to take the derivative of \( \cos(3y)\) while \( \sin(3x)\) remains constant, resulting in \( - 3 \sin(3x) \sin(3y)\).

Trigonometric functions like \( \sin \), \( \cos \), and \( \tan \) come with their own set of rules for differentiation. The derivatives of the basic trigonometric functions are usually memorized:

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

When you differentiate trigonometric functions that are functions of another function, as in \( \sin 3x \) or \( \cos 3y \), you need to apply the chain rule to account for the inner function. For instance, the derivative of \( \sin 3x \) with respect to \( x \) is \( 3\cos 3x \), because while the derivative of \( \sin \) is \( \cos \), you also multiply by the derivative of the inner function 3x, which is 3. This concept is pivotal for understanding the steps involved in solving the original textbook problem, ensuring we arrive at \( \frac{\partial z}{\partial x} = 3 \cos(3x) \cos(3y) \) and \( \frac{\partial z}{\partial y} = - 3 \sin(3x) \sin(3y) \).