Problem 120 - Calculus Volume 3

Question

For the following exercises, calculate the partial derivatives. \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for \(z=x^{8} e^{3 y}\).

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial x} = 8x^7 e^{3y}\) and \(\frac{\partial z}{\partial y} = 3x^8 e^{3y}\).

Step 1: Identify the Function

The function given in the problem is \(z = x^8 e^{3y}\). This function is composed of two variables: \(x\) and \(y\). Our task is to find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

Step 2: Find \(\frac{\partial z}{\partial x}\)

To calculate the partial derivative of \(z\) with respect to \(x\), we treat \(y\) as a constant. The derivative of \(x^8\) with respect to \(x\) is \(8x^7\), and the exponential part, \(e^{3y}\), stays constant as it is independent of \(x\). Therefore, \[\frac{\partial z}{\partial x} = 8x^7 e^{3y}.\]

Step 3: Find \(\frac{\partial z}{\partial y}\)

Now, calculate the partial derivative with respect to \(y\), treating \(x\) as a constant. The derivative of \(e^{3y}\) with respect to \(y\) is \(3e^{3y}\), and \(x^8\) remains constant. Therefore, \[\frac{\partial z}{\partial y} = x^8 \cdot 3 e^{3y} = 3x^8 e^{3y}.\]

Key Concepts

Multivariable Calculus

Multivariable calculus expands calculus concepts to functions involving more than one variable, such as those found in physics and engineering. In mathematics, many real-world problems involve several variables interacting with each other. Thus, understanding how functions change with respect to multiple inputs is crucial. Unlike single-variable calculus, where functions have only one input variable, multivariable calculus deals with functions like \( z = f(x, y) \), which have two or more inputs. It's essential because:

• It allows us to study surfaces and curves in higher dimensions.
• It provides tools for optimization when several parameters influence outcomes.
• It helps in modeling complex systems in sciences and engineering.

To grasp multivariable calculus, one should be comfortable with visualizing and analyzing functions that depend on more than one variable.

Differentiation with respect to variables

Differentiation with respect to variables in a multivariable function involves finding partial derivatives. A partial derivative measures how a function changes as only one of the variables changes, while keeping all other variables constant.For example, given a function \( z = x^8 e^{3y} \), we have two variables: \( x \) and \( y \). Calculating partial derivatives here means:

• For \( \frac{\partial z}{\partial x} \), consider \( y \) constant and differentiate with respect to \( x \).
• For \( \frac{\partial z}{\partial y} \), treat \( x \) as constant and differentiate with respect to \( y \).

This kind of differentiation helps isolate the effect each variable has on the function, a foundational concept in analyzing multivariable systems.

Chain Rule in Calculus

The chain rule is a powerful tool in calculus for differentiating composite functions. When functions are composed of other functions, the chain rule provides a way to find the derivative by multiplying the derivatives of the inner and outer functions.In the context of partial derivatives and multivariable calculus, the chain rule helps when variables are interdependent or when functions are expressed in terms of other functions. Consider a scenario where \( z = g(u(x, y), v(x, y)) \). Here,

• The derivative of \( u \) with respect to \( x \) affects \( z \).
• The derivative of \( v \) with respect to \( y \) also influences \( z \).

Using the chain rule in multivariable calculus can simplify calculations dramatically by accounting for how changes in intermediate functions affect the final function. This rule is indispensable in fields such as physics and engineering, where complex systems are modeled by multivariable functions.