Problem 118 - Calculus 3

Question

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

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial x} = 3\cos(3x)\cos(3y)\)

Step 1: Differentiate the inner function with respect to x

The given function is a product of two functions: \(\sin(3x)\) and \(\cos(3y)\). The derivative of \(\sin(3x)\) with respect to \(x\) is \(3\cos(3x)\) because of the chain rule, where the derivative of \(3x\) with respect to \(x\) is 3.

Step 2: Apply the product rule

Since \(z=\sin(3x)\cos(3y)\) is a product of two functions of different variables, apply the product rule: \(\frac{d}{dx}[u(x)\cdot v(x)] = u'(x)v(x) + u(x)v'(x)\). Here, \(u(x) = \sin(3x)\) and \(v(y) = \cos(3y)\).

Step 3: Differentiate (Using Product Rule)

For \(\frac{\partial z}{\partial x}\), only \(\sin(3x)\) depends on \(x\), hence \(v'(y)=0\). Applying the product rule:\[ \frac{\partial z}{\partial x} = \frac{d}{dx}[\sin(3x)]\cdot \cos(3y) = 3\cos(3x)\cos(3y) \]

Key Concepts

Chain Rule

Partial derivatives often involve functions within functions. This is where the chain rule comes into play. The chain rule assists in differentiation when a composite function is present. Suppose you have a function like \( f(g(x)) \). The chain rule lets you differentiate this by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

• First, identify the inner and outer functions. In the example given, \( g(x) = 3x \) is the inner function, while \( \sin(u) \) represents the outer function.
• Differentiate the outer function with respect to the inner function, while ignoring the inner function first.
• Next, multiply by the derivative of the inner function.

In the exercise solution, \( \sin(3x) \) required the chain rule. The derivative of \( \sin(u) \) is \( \cos(u) \), and when plugged into the chain rule, it yields \( 3\cos(3x) \). This is because \( g(x) = 3x\) has a derivative of 3. Using these steps, you can tackle many types of composite functions through differentiation.

Product Rule

When you have a product of two functions, \( u(x) \) and \( v(x) \), and you want to differentiate, that's where the product rule steps in. It's essential when dealing with products, as it allows you to differentiate each part separately and then combine the results.

• The product rule is given by: \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) \).
• Start by differentiating each function on its own. The functions can vary by variable, as seen in the example \( \sin(3x) \cdot \cos(3y) \).
• Apply the derivatives to the rule's formula.

Since only \( \sin(3x) \) depends on \( x \), the derivative of \( \cos(3y) \) with respect to \( x \) is zero. Therefore, \( u'(x) = 3\cos(3x) \) becomes the only part needing differentiation while \( \cos(3y) \) stays constant allowing it to multiply.

Differentiation

Differentiation is a fundamental process in calculus used to find rates of change. When working with partial derivatives, differentiation involves considering how functions change in one direction, keeping other variables constant.

• Partial differentiation implies fixing variables outside the derivative and differentiating with respect to one variable.
• The focus is on how a multiplicative component like \( \sin(3x) \) changes when \( x \) is varied, without affecting \( \cos(3y) \).
• It’s important to identify which variables are independent or constant during the differentiation process in partial derivatives.

For example, when you perform \( \frac{\partial z}{\partial x} \) for \( z=\sin (3 x) \cos (3 y) \), you're essentially looking at how \( z \) changes with respect to \( x \), treating \( y \) as a constant. Mastering this strategy for partial differentiation gives powerful insight into multidimensional functions.