Problem 31 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$w=\ln (x+2 y+3 z)$$

Step-by-Step Solution

Answer

The partial derivatives are \(\frac{1}{x+2y+3z}\), \(\frac{2}{x+2y+3z}\), \(\frac{3}{x+2y+3z}\) for \(x\), \(y\), \(z\) respectively.

Step 1: Identify the Function

We are given the function to consider, which is \( w = \ln(x + 2y + 3z) \). The aim is to find the first partial derivatives with respect to \(x\), \(y\), and \(z\).

Step 2: Partial Derivative with Respect to x

To find the partial derivative of \(w\) with respect to \(x\), treat \(y\) and \(z\) as constants. The derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\), so apply the chain rule:\[ \frac{\partial w}{\partial x} = \frac{1}{x + 2y + 3z} \cdot \frac{\partial}{\partial x}(x + 2y + 3z) = \frac{1}{x + 2y + 3z} \cdot 1 = \frac{1}{x + 2y + 3z}. \]

Step 3: Partial Derivative with Respect to y

To find the partial derivative of \(w\) with respect to \(y\), treat \(x\) and \(z\) as constants. The derivative is:\[ \frac{\partial w}{\partial y} = \frac{1}{x + 2y + 3z} \cdot \frac{\partial}{\partial y}(x + 2y + 3z) = \frac{1}{x + 2y + 3z} \cdot 2 = \frac{2}{x + 2y + 3z}. \]

Step 4: Partial Derivative with Respect to z

To find the partial derivative of \(w\) with respect to \(z\), treat \(x\) and \(y\) as constants. The derivative is:\[ \frac{\partial w}{\partial z} = \frac{1}{x + 2y + 3z} \cdot \frac{\partial}{\partial z}(x + 2y + 3z) = \frac{1}{x + 2y + 3z} \cdot 3 = \frac{3}{x + 2y + 3z}. \]

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with rates of change and accumulation. At its core, calculus has two fundamental concepts: derivatives and integrals.
Derivatives measure how a function changes as its input changes, and integrals are about accumulating quantities. In this exercise, we are specifically interested in the idea of derivatives. Understanding derivatives can help us know how a function's output changes in response to changes in its input. This is crucial for solving problems involving rates of change, such as how quickly a car accelerates or how the economy grows. In a sense, derivatives give us a way to "zoom in" and see what a function is doing instantly at any particular point.
When we extend the idea of a derivative to functions with more than one variable, such as the one seen in the exercise, we enter the realm of partial derivatives.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate compositions of functions. When you have a function inside another function, the chain rule helps you find the derivative of the outer function in terms of the inner function.
For example, if you want to differentiate a composite function like \( f(g(x)) \), the chain rule tells us that the derivative is the product of the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function itself.In the context of partial derivatives, the chain rule is used to systematically break down and compute derivatives with respect to one variable while keeping others constant.

• Think of it as a way to navigate through a chain of functions, moving from one function to the next, calculating derivatives step by step.
• The result we apply to solve this exercise is a direct result of using the chain rule to differentiate the natural logarithm function, \( \ln(x + 2y + 3z) \).

It simplifies the process by letting us separate the computation into manageable parts.

Multivariable Calculus

Multivariable calculus extends the ideas of calculus to functions with several variables. This area of calculus enables us to work with and understand systems where change is happening in more than one direction simultaneously.
In functions like \( w = \ln(x + 2y + 3z) \), you have three variables: \(x\), \(y\), and \(z\). Each partial derivative represents the rate of change of the function as one specific variable changes, while others are held constant.Partial derivatives are akin to taking a slice of the multivariable function to see how change unfolds with respect to one variable.

• When computing \( \frac{\partial w}{\partial x} \), you treat \( y \) and \( z \) as constants, isolating the effect of \( x \).
• Similarlly, \( \frac{\partial w}{\partial y} \) and \( \frac{\partial w}{\partial z} \) provide insights on changes with respect to \( y \) and \( z \).

Mastering multivariable calculus is vital, especially in fields that deal with complex systems like physics or engineering, as it helps model and solve real-world problems that depend on several variables.