Problem 34 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$u=X^{y / z}$$

Step-by-Step Solution

Answer

The first partial derivatives are: \( \partial u/\partial X = \frac{y}{z}X^{(y/z) - 1} \), \( \partial u/\partial y = \frac{1}{z}X^{y/z} \ln(X) \), \( \partial u/\partial z = -\frac{y}{z^2}X^{y/z} \ln(X) \).

Step 1: Understand the Function

We are given the function \( u = X^{y/z} \). This function involves three variables: \( X \), \( y \), and \( z \). To find the first partial derivatives, we need to differentiate the function with respect to each variable separately, treating the other variables as constants.

Step 2: Differentiate with Respect to X

To find the partial derivative of \( u \) with respect to \( X \), treat \( y \) and \( z \) as constants. The derivative of \( X^{n} \) with respect to \( X \) is \( nX^{n-1} \). Apply this to our function.\[ \frac{\partial u}{\partial X} = \frac{y}{z}X^{(y/z) - 1}. \]

Step 3: Differentiate with Respect to y

To find the partial derivative of \( u \) with respect to \( y \), treat \( X \) and \( z \) as constants. Recall that the derivative of \( a^{bx} \) with respect to \( x \) is \( a^{bx} \ln(a)b \). Using the chain rule, we have:\[ \frac{\partial u}{\partial y} = X^{y/z} \cdot \ln(X) \cdot \frac{1}{z} = \frac{1}{z} X^{y/z} \ln(X). \]

Step 4: Differentiate with Respect to z

To find the partial derivative of \( u \) with respect to \( z \), treat \( X \) and \( y \) as constants. We apply the derivative of the function \( a^{bx/c} \) with respect to \( c \) using the chain rule:\[ \frac{\partial u}{\partial z} = -\frac{y}{z^2} X^{y/z} \ln(X). \]

Key Concepts

Calculus

Calculus plays a crucial role in understanding and solving problems involving rates of change and areas under curves. It consists of two main branches: differential calculus and integral calculus.
While integral calculus deals with accumulation and areas, differential calculus focuses on the concept of rates of change. In our exercise, we focus on differential calculus as we deal with derivatives.
Derivatives allow us to see how a function changes as its input changes. This is vital in science and engineering, where we often need to predict how a system behaves under changing conditions.

Multivariable Functions

In calculus, a function is typically a relation between a set of inputs and a single output. Multivariable functions extend this by involving more than one input variable.
Our function, specifically, involves three variables: \(X\), \(y\), and \(z\). In multivariable calculus, partial derivatives are used to measure how a function changes as one of its variables changes, holding the others constant.

• The function \(u = X^{y/z}\) requires us to consider how \(u\) changes with respect to \(X\), \(y\), and \(z\) separately.
• This is why partial derivatives are essential when dealing with multivariable functions, to isolate and understand the effect of each variable on the function.

Understanding multivariable functions is critical as they are widely used in fields like physics, engineering, and economics.

Differentiation

Differentiation is the process of finding the derivative of a function, which tells us how the function changes as its input changes. In multivariable calculus, we use partial differentiation.
Partial differentiation involves differentiating a multivariable function with respect to one variable while keeping the others constant.

• For example, to find \(\frac{\partial u}{\partial X}\) for \(u = X^{y/z}\), we treat \(y\) and \(z\) as constants.
• This helps determine the rate at which \(u\) changes solely with respect to \(X\).

This technique is integral to many applications, such as optimization problems, where one seeks to determine how changes in one variable affect the overall system.

Chain Rule

The chain rule is a vital tool in calculus for differentiating composite functions. It allows us to find the derivative of a function that is composed of other functions.

• In our exercise, we see its application when differentiating with respect to \(y\) and \(z\).
• The chain rule tells us that if we have a function \(f(g(x))\), its derivative can be found using \(f'(g(x)) \cdot g'(x)\).

When applying the chain rule:

• When differentiating \(u\) with respect to \(y\), \(X^{y/z} \cdot \ln(X) \cdot \frac{1}{z}\) is the result, showcasing the power of the chain rule to handle complex combinations.
• This is especially useful for working with exponential and logarithmic functions in multivariable contexts.

Mastering the chain rule is crucial for effectively handling more complicated operations in calculus.