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Chapter 13: Problem 15

Find both first partial derivatives. $$ z=\ln \frac{x+y}{x-y} $$

Short Answer

Expert verified
Thus the first partial derivatives of the function \(\ln \frac{x+y}{x-y}\) are: \(\frac{\partial z}{\partial x} = -2y/(z*(x - y)^2)\) and \(\frac{\partial z}{\partial y} = 2x/(z*(x - y)^2)\).

Step by step solution

01

Compute the partial derivative with respect to x

Start by finding the derivative with respect to \(x\). Use the chain rule which states that \(d(f(g(x))/dx = f'(g(x)).g'(x)\). Apply logarithmic differentiation. First, differentiate the outer function to get \(1/z\) and then multiply by the derivative of the inner function. The derivative of the inner function is calculated by using the quotient rule for differentiation which is \((f/g)' = (f'g - g'f)/g^2\). The functions \(f(x) = x + y\) and \(g(x) = x - y\). Finding their derivatives with respect to \(x\) gives \(f'(x) = 1\) and \(g'(x) = 1\). Substituting these values into the quotient rule gives \((f/g)' = (1*(x - y) - (x + y)*1)/(x - y)^2 = -2*y/(x - y)^2\). So the derivative of the function \(\ln (x+y)/(x-y) \) with respect to \(x\) is: \(\frac{\partial z}{\partial x} = (1/z) (-2y/(x - y)^2) = -2y/(z*(x - y)^2)\).
02

Compute the partial derivative with respect to y

The process is similar to step 1, except now the derivatives of \(f(x) = x + y\) and \(g(x) = x - y\) are taken with respect to \(y\), not \(x\). Thus \(f'(y) = 1\) and \(g'(y) = -1\). Substituting these values into the quotient rule gives \((f/g)' = (1*(x - y) - (x + y)*(-1))/(x - y)^2 = 2x/(x - y)^2\). This gives the derivative of the function \(\ln (x+y)/(x-y) \) with respect to \(y\): \(\frac{\partial z}{\partial y} = (1/z) (2x/(x - y)^2) = 2x/(z*(x - y)^2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule in Partial Differentiation
Understanding the chain rule is critical when dealing with complex functions, particularly in the realm of partial differentiation. It's a principle that allows us to differentiate composite functions, meaning functions that can be broken down into two or more simpler functions. The chain rule states that the derivative of a composite function is the derivative of the outer function with respect to the inner function multiplied by the derivative of the inner function with respect to its variable.

When applying the chain rule to find the partial derivative of a function like \( z = \ln\frac{x+y}{x-y} \), one must look at the 'inner function'鈥攊n this case, \( \frac{x+y}{x-y} \)鈥攁nd the 'outer function'鈥攚hich is the natural logarithm, \( \ln(x) \). Differentiating the outer function gives us \( \frac{1}{u} \) (where \( u \) is the inner function), which we then multiply by the derivative of the inner function to obtain the full derivative with respect to either variable.
Quotient Rule for Differentiation
The quotient rule is an essential tool when you encounter a function that's a ratio of two differentiable functions. It allows us to find the derivative of a function that is divided by another. The quotient rule formula is given by \( (\frac{f}{g})' = \frac{f'g - g'f}{g^2} \), where \( f \) and \( g \) are functions of the same variable and \( g \eq 0\).

When applying the quotient rule to differentiate \( \frac{x+y}{x-y} \), we recognize \( f(x) = x + y \) and \( g(x) = x - y \) as the top and bottom functions of our ratio. Differentiating each with respect to \( x \) or \( y \) separately gives us the numerators for our quotient rule formula. Applying this step-by-step, as seen in the exercise, ensures a clear path to obtaining the partial derivatives needed.
Logarithmic Differentiation
Special consideration must be given to logarithmic differentiation, which is a technique used when differentiating logarithmic functions, especially when the argument of the logarithm is a complex function itself. This method simplifies the differentiation process since logarithms transform multiplication into addition, division into subtraction, and powers into products.

In the context of our exercise, after applying the chain rule, logarithmic differentiation comes into play because we're tasked with differentiating \( \ln\frac{x+y}{x-y} \). The derivative of the natural logarithm \( \ln(u) \) is \( \frac{1}{u} \) times the derivative of \( u \) with respect to its variable. This final step is what allows us to compute the partial derivatives of the complex logarithmic function provided in the original problem.

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Most popular questions from this chapter

A cargo container (in the shape of a rectangular solid) must have a volume of 480 cubic feet. The bottom will cost \(\$ 5\) per square foot to construct and the sides and the top will cost \(\$ 3\) per square foot to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has minimum cost.

Find the least squares regression line for the points. Use the regression capabilities of a graphing utility to verify your results. Use the graphing utility to plot the points and graph the regression line. $$ (1,0),(3,3),(5,6) $$

Find the point on the surface where the tangent plane is horizontal. Use a computer algebra system to graph the surface and the horizontal tangent plane. Describe the surface where the tangent plane is horizontal.\(z=3 x^{2}+2 y^{2}-3 x+4 y-5\)

Use Lagrange multipliers to find the indicated extrema, assuming that \(x, y\), and \(z\) are positive. Maximize \(f(x, y, z)=x y z\) Constraint: \(x+y+z-6=0\)

(a) Use Lagrange multipliers to prove that the product of three positive numbers \(x, y\), and \(z\), whose sum has the constant value \(S\), is a maximum when the three numbers are equal. Use this result to prove that \(\sqrt[3]{x y z} \leq \frac{x+y+z}{3}\). (b) Generalize the result of part (a) to prove that the product \(x_{1} x_{2} x_{3} \cdot \cdots x_{n}\) is a maximum when \(x_{1}=x_{2}=x_{3}=\) \(\cdots=x_{n}, \sum_{i=1}^{n} x_{i}=S\), and all \(x_{i} \geq 0 .\) Then prove that $$ \sqrt[n]{x_{1} x_{2} x_{3} \cdot \cdots x_{n}} \leq \frac{x_{1}+x_{2}+x_{3}+\cdots \cdot+x_{n}}{n}. $$ This shows that the geometric mean is never greater than the arithmetic mean.
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