Problem 27
Question
Find the first partial derivatives of the given function. $$ g(u, v)=\ln \left(4 u^{2}+5 v^{3}\right) $$
Step-by-Step Solution
Verified Answer
The partial derivatives are \( \frac{8u}{4u^2 + 5v^3} \) and \( \frac{15v^2}{4u^2 + 5v^3} \).
1Step 1: Identify the Function and Variables
The given function is \( g(u, v) = \ln(4u^2 + 5v^3) \). We are tasked with finding the partial derivatives with respect to \( u \) and \( v \). This means we need to compute \( \frac{\partial g}{\partial u} \) and \( \frac{\partial g}{\partial v} \).
2Step 2: Apply the Chain Rule
To find the partial derivative, we first note that the function has the form \( \ln(f(u, v)) \). The derivative of \( \ln(x) \) is \( \frac{1}{x} \), so using the chain rule, \( \frac{\partial}{\partial u}\ln(f(u,v)) = \frac{1}{f(u,v)} \cdot \frac{\partial f}{\partial u} \).
3Step 3: Differentiate with Respect to \( u \)
First, find \( \frac{\partial}{\partial u}(4u^2 + 5v^3) \). As \(5v^3\) is a constant when differentiating with respect to \( u \), the derivative is \(8u\). Thus, \(\frac{\partial g}{\partial u} = \frac{1}{4u^2 + 5v^3} \cdot 8u = \frac{8u}{4u^2 + 5v^3} \).
4Step 4: Differentiate with Respect to \( v \)
Similarly, find \( \frac{\partial}{\partial v}(4u^2 + 5v^3) \). Here \( 4u^2 \) is constant with respect to \( v \), so the derivative is \(15v^2\). Thus, \(\frac{\partial g}{\partial v} = \frac{1}{4u^2 + 5v^3} \cdot 15v^2 = \frac{15v^2}{4u^2 + 5v^3} \).
5Step 5: Summary of Partial Derivatives
We have found the partial derivatives: \( \frac{\partial g}{\partial u} = \frac{8u}{4u^2 + 5v^3} \) and \( \frac{\partial g}{\partial v} = \frac{15v^2}{4u^2 + 5v^3} \). These represent the rates of change of \( g \) with respect to \( u \) and \( v \), respectively.
Key Concepts
Chain RuleNatural LogarithmDifferentiationMultivariable Calculus
Chain Rule
The chain rule is an essential concept in calculus, especially when dealing with composite functions. A composite function occurs when you have a function inside another function, for example, \( g(u, v) = \ln(4u^2 + 5v^3) \). The chain rule helps us differentiate such functions by allowing us to compute the derivative of the outer function and multiply it by the derivative of the inner function.
In our exercise, the outer function is the natural logarithm, and the inner function is \( f(u, v) = 4u^2 + 5v^3 \). When applying the chain rule, start by taking the derivative of the natural logarithm, which is \( \frac{1}{f(u, v)} \). Then, multiply it by the derivative of the inner function concerning the variable of interest.
This step-by-step approach simplifies the complex process of finding partial derivatives in multivariable functions, making it a powerful tool in multivariable calculus.
In our exercise, the outer function is the natural logarithm, and the inner function is \( f(u, v) = 4u^2 + 5v^3 \). When applying the chain rule, start by taking the derivative of the natural logarithm, which is \( \frac{1}{f(u, v)} \). Then, multiply it by the derivative of the inner function concerning the variable of interest.
This step-by-step approach simplifies the complex process of finding partial derivatives in multivariable functions, making it a powerful tool in multivariable calculus.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. It is an important mathematical function because it describes many natural phenomena.
For differentiation, the derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \). This derivative is fundamental when dealing with logarithmic functions, such as in our exercise
where the function \( g(u, v) \) is expressed as \( \ln(4u^2 + 5v^3) \). Understanding how the natural logarithm works is crucial because it frequently appears in multivariable calculus, especially in problems involving growth and decay.
For differentiation, the derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \). This derivative is fundamental when dealing with logarithmic functions, such as in our exercise
where the function \( g(u, v) \) is expressed as \( \ln(4u^2 + 5v^3) \). Understanding how the natural logarithm works is crucial because it frequently appears in multivariable calculus, especially in problems involving growth and decay.
Differentiation
Differentiation is the process of finding a derivative, which signifies the rate of change of a function concerning one of its variables. In univariate functions, it involves basic calculus rules applicable to simple functions.
However, in multivariable functions, such as \( g(u, v) = \ln(4u^2 + 5v^3) \), the process requires the use of partial derivatives. A partial derivative is the derivative with respect to one variable while keeping other variables constant.
This is especially important when we encounter functions dependent on multiple variables. Differentiating with respect to \( u \), we're treating \( v \) as a constant and vice versa. This nuanced differentiation is pivotal in understanding more complex calculations and analyses in multivariable calculus.
However, in multivariable functions, such as \( g(u, v) = \ln(4u^2 + 5v^3) \), the process requires the use of partial derivatives. A partial derivative is the derivative with respect to one variable while keeping other variables constant.
This is especially important when we encounter functions dependent on multiple variables. Differentiating with respect to \( u \), we're treating \( v \) as a constant and vice versa. This nuanced differentiation is pivotal in understanding more complex calculations and analyses in multivariable calculus.
Multivariable Calculus
Multivariable calculus expands upon the concepts of single-variable calculus by dealing with functions of more than one variable. It introduces new concepts like partial derivatives, multiple integrals, and vector calculus.
In our problem, the function \( g(u, v) = \ln(4u^2 + 5v^3) \) is dependent on two variables, \( u \) and \( v \). Multivariable calculus allows us to understand how this function changes concerning each variable separately and together.
In our problem, the function \( g(u, v) = \ln(4u^2 + 5v^3) \) is dependent on two variables, \( u \) and \( v \). Multivariable calculus allows us to understand how this function changes concerning each variable separately and together.
- Partial Derivatives: These measure how the function changes when one variable changes, with others held constant.
- Applications: Used in fields like physics, engineering, and economics, where systems depend on multiple factors.
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