Problem 119 - Calculus 3

Question

For the following exercises, calculate the partial derivatives. \(\frac{\partial z}{\partial y}\) for \(z=\sin (3 x) \cos (3 y)\)

Step-by-Step Solution

Answer

The partial derivative is \( -3 \sin(3x) \sin(3y) \).

Step 1: Understand the Function

We are given the function \(z = \sin(3x) \cos(3y)\). We need to find the partial derivative of \(z\) with respect to \(y\). This means treating \(x\) as a constant while differentiating with respect to \(y\).

Step 2: Apply the Product Rule

The function \(z = \sin(3x) \cos(3y)\) is a product of two functions of \(y\). For \(z = u(y)v(y)\), the partial derivative is \(\frac{\partial z}{\partial y} = u'(y)v(y) + u(y)v'(y)\). Here, \(u(y) = \sin(3x)\) which becomes constant, and \(v(y) = \cos(3y)\).

Step 3: Differentiate \(v(y) = \cos(3y)\)

Differentiate \(\cos(3y)\) with respect to \(y\): \( \frac{d}{dy} \cos(3y) = -3 \sin(3y) \). This occurs because the derivative of \(\cos(kx)\) is \(-k \sin(kx)\).

Step 4: Calculate the Partial Derivative

Plug the derivatives into the product rule: \( \frac{\partial z}{\partial y} = \sin(3x)(-3 \sin(3y)) \). Simplifying gives \( \frac{\partial z}{\partial y} = -3 \sin(3x) \sin(3y) \).

Key Concepts

Product Rule

The product rule is a handy formula used in calculus to find the derivative of a product of two functions. When you have two functions, say \(u(y)\) and \(v(y)\), their product is \(z = u(y) \cdot v(y)\). The derivative of this product with respect to \(y\) is found using the product rule, given by:

• \(\frac{d}{dy}[u(y) \cdot v(y)] = u'(y)v(y) + u(y)v'(y)\)

This means we take the derivative of the first function, \(u(y)\), times the second function, \(v(y)\), plus the first function, \(u(y)\), times the derivative of the second function, \(v(y)\). It's like treating each function separately and then combining results using this formula.

In the context of partial differentiation, if one of the functions doesn't actually depend on \(y\), it simplifies the process, as seen in our problem where \(u(y) = \sin(3x)\), which is constant with respect to \(y\). This makes using the product rule straightforward because it reduces to differentiating only the non-constant term.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes. In the context of partial derivatives, we consider how a function changes with respect to one variable while keeping other variables constant.

In our problem, we differentiated with respect to \(y\). Since \(z = \sin(3x) \cos(3y)\), we consider \(x\) as a constant, allowing us to focus on how \(z\) changes with \(y\). The derivative becomes a measure of this rate of change focusing specifically on the impact of \(y\).

Differentiation involves applying rules such as the chain rule and product rule, depending on the structure of the function. For our function, after identifying \(v(y) = \cos(3y)\), differentiation yields \(-3 \sin(3y)\), using the chain rule within the differentiation of the cosine function, indicating the rate of change of the trigonometric component with respect to \(y\).

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus, especially in problems involving oscillatory functions. Understanding their derivatives is key to solving calculus exercises.

Consider the function \(z = \sin(3x) \cos(3y)\). The terms involve basic trigonometric functions \(\sin\) and \(\cos\), which have well-defined derivatives:

• The derivative of \(\sin(kx)\) is \(k \cos(kx)\).
• The derivative of \(\cos(kx)\) is \(-k \sin(kx)\).

These rules are derived from the basic definitions of sine and cosine and their behavior under differentiation, known as the chain rule, which considers the constant multiplier inside the argument.

For our derivative \(\frac{d}{dy} \cos(3y)\), we applied this knowledge to find \(-3 \sin(3y)\). Recognizing how these transformations work helps greatly in differentiating such functions, forming a basis for more complex work with trigonometric functions in calculus.