Problem 47
Question
Find the indicated partial derivatives. \(f(x, y)=\ln (x+y) ; \frac{\partial^{2} f}{\partial x^{2}}\)
Step-by-Step Solution
Verified Answer
\( \frac{\partial^2 f}{\partial x^2} = -\frac{1}{(x+y)^2} \).
1Step 1: Understanding the Function
We need to find the second partial derivative of the function \( f(x, y) = \ln(x+y) \) with respect to \( x \). This involves differentiating the function twice concerning \( x \).
2Step 1: First Partial Derivative with Respect to x
First, we compute \( \frac{\partial f}{\partial x} \). Using the chain rule, the derivative of \( \ln(x+y) \) with respect to \( x \) is \( \frac{1}{x+y} \). Thus, \( \frac{\partial f}{\partial x} = \frac{1}{x+y} \).
3Step 2: Second Partial Derivative with Respect to x
Now, differentiate \( \frac{\partial f}{\partial x} = \frac{1}{x+y} \) again with respect to \( x \). The derivative of \( \frac{1}{x+y} \) is obtained by treating \( y \) as a constant: \( -\frac{1}{(x+y)^2} \). Therefore, \( \frac{\partial^2 f}{\partial x^2} = -\frac{1}{(x+y)^2} \).
Key Concepts
Second Partial DerivativeChain RuleLogarithmic Functions
Second Partial Derivative
The second partial derivative gives us insight into how a function's rate of change itself changes in relation to a specific variable, typically providing details about the concavity or the curvature in that direction. When calculating the second partial derivative of a function like \( f(x, y) = \ln(x+y) \), we first need to determine the first derivative with respect to \( x \). This involves holding other variables constant and differentiating once.
- First Derivative: The first partial derivative \( \frac{\partial f}{\partial x} \) is \( \frac{1}{x+y} \).
- Second Derivative: To find the second partial derivative, we differentiate \( \frac{1}{x+y} \) with respect to \( x \) again, resulting in \( -\frac{1}{(x+y)^2} \).
Chain Rule
The chain rule is a fundamental technique in calculus that allows us to differentiate composite functions effectively. When dealing with partial derivatives, the chain rule becomes particularly useful for functions composed of other functions, like \( \ln(x+y) \). Here, the chain rule helps us differentiate when variables inside functions are combined.
- Function Inside Another: In \( \ln(x+y) \), the expression \( x+y \) is considered as one composite function within the natural logarithm \( \ln \).
- Applying Chain Rule: When differentiating \( \ln(x+y) \) with respect to \( x \), we treat \( y \) as a constant. The chain rule means we first find the derivative of the outer function (natural logarithm), then multiply by the derivative of the inner function \( x+y \).
- Result: This gives us the partial derivative \( \frac{1}{x+y} \), simplifying our computation when advancing to second derivatives.
Logarithmic Functions
Logarithmic functions, especially the natural logarithm \( \ln \), play a crucial role in mathematical analysis and calculus. These functions are the inverse of exponential functions, which are widespread across sciences for modeling growth processes.
- Natural Logarithm \( \ln \): It's a logarithm to the base \( e \), where \( e \approx 2.718 \). This is frequently used in calculus due to its simple differentiation rules: the derivative of \( \ln(u) \) is \( \frac{1}{u} \).
- In Partial Derivatives: For functions like \( \ln(x+y) \), taking the derivative transforms these logarithmic expressions into fractions, which are more straightforward to handle, especially when using rules like the chain rule.
- Behavior Analysis: Logarithms are useful in understanding the growth and behavior of functions. They grow slowly and are defined for positive values of their argument, making them ideal for scaling or compressing large datasets in analysis.
Other exercises in this chapter
Problem 46
Find a linear approximation to $$ \mathbf{f}(x, y)=\left[\begin{array}{c} \sqrt{2 x+y} \\ x-y^{2} \end{array}\right] $$ at \((1,2) .\) Use your result to find a
View solution Problem 46
Let $$f(x, y)=x+y \quad(x, y) \in \mathbf{R}^{2}$$ with constraint function \(x y=1\). (a) Use Lagrange multipliers to find all local extrema. (b) Are there glo
View solution Problem 47
The Lotka-Volterra equations are often used to model the links between a particular of prey organisms (e.g., sardines) and a population of predatory organisms (
View solution Problem 47
Let $$f(x, y)=x+y$$ with constraint function $$\frac{1}{x}+\frac{1}{y}=1, x \neq 0, y \neq 0$$ (a) Use Lagrange multipliers to find all local extrema. (b) Are t
View solution