Chapter 13: Problem 17
Find \(\partial z / \partial x\) and \(\partial z / \partial y\) $$ z=4 e^{x^{2} y^{3}} $$
Short Answer
Expert verified
\( \frac{\partial z}{\partial x} = 8xy^3 e^{x^2y^3} \) and \( \frac{\partial z}{\partial y} = 12x^2y^2 e^{x^2y^3} \).
Step by step solution
01
Function Definition
Firstly, we identify the function that we need to differentiate. The function given is: \[ z = 4e^{x^2y^3} \] This will be differentiated with respect to the variables \(x\) and \(y\).
02
Differentiate with respect to x
To find \( \frac{\partial z}{\partial x} \), apply the chain rule for differentiation. Notice that \(x^2y^3\) is the exponent of the exponential function:1. Identify the inner function as \( u = x^2y^3 \).2. The derivative of \( e^u \) with respect to \( u \) is \( e^u \).3. Find \( \frac{\partial u}{\partial x} = 2xy^3 \).4. Apply the chain rule: \[ \frac{\partial z}{\partial x} = 4e^{x^2y^3} \cdot \frac{\partial u}{\partial x} = 4e^{x^2y^3} \cdot (2xy^3) \]5. Simplify: \[ \frac{\partial z}{\partial x} = 8xy^3 e^{x^2y^3} \]
03
Differentiate with respect to y
To find \( \frac{\partial z}{\partial y} \), again apply the chain rule:1. Identify the inner function again as \( u = x^2y^3 \).2. The derivative of \( e^u \) with respect to \( u \) is \( e^u \).3. Find \( \frac{\partial u}{\partial y} = 3x^2y^2 \).4. Apply the chain rule: \[ \frac{\partial z}{\partial y} = 4e^{x^2y^3} \cdot \frac{\partial u}{\partial y} = 4e^{x^2y^3} \cdot (3x^2y^2) \]5. Simplify: \[ \frac{\partial z}{\partial y} = 12x^2y^2 e^{x^2y^3} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental concept in calculus, particularly useful when dealing with functions composed of other functions. Here, we focus on applying it to partial derivatives.
When you have a function like \[ z = 4e^{x^2y^3} \]understand that the exponent, \( x^2y^3 \), is itself a function of both \( x \) and \( y \). This is what we call the "inner function." The chain rule tells us how to differentiate such composite functions.
To apply the chain rule:
When you have a function like \[ z = 4e^{x^2y^3} \]understand that the exponent, \( x^2y^3 \), is itself a function of both \( x \) and \( y \). This is what we call the "inner function." The chain rule tells us how to differentiate such composite functions.
To apply the chain rule:
- Identify the inner function. Here, it is \( u = x^2y^3 \).
- Recognize that the outer function is an exponential, \( e^u \).
- Differentiate the outer function with respect to the inner function. For an exponential function, the derivative of \( e^u \) is simply \( e^u \).
- Find the derivative of the inner function, \( u \), with respect to each variable you are interested in.
- Multiply these derivatives together according to the chain rule.
Exponential Functions
Exponential functions are a critical topic in calculus and have many applications in science and engineering. The natural exponential function \( e^x \) is particularly important due to its unique properties.
The function given, \[ z = 4e^{x^2y^3} \]is an exponential function with a complex exponent. Exponential functions are unique because:
The function given, \[ z = 4e^{x^2y^3} \]is an exponential function with a complex exponent. Exponential functions are unique because:
- Their rate of growth is proportional to their current value. This means \( e^u \) grows faster as \( u \) increases.
- The derivative of the exponential function \( e^u \) is the function itself, \( e^u \), which simplifies the differentiation process.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions with more than one variable. This means instead of just finding how a function changes with one variable, we examine how it changes with respect to multiple variables, separately and together.
The expression \[ z = 4e^{x^2y^3} \]embodies multivariable calculus because it includes both \( x \) and \( y \). When calculating partial derivatives like \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), you:
The expression \[ z = 4e^{x^2y^3} \]embodies multivariable calculus because it includes both \( x \) and \( y \). When calculating partial derivatives like \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), you:
- Focus on changing one variable at a time, treating others as constants.
- Use techniques such as the chain rule tailored for multiple variables.