/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Find \(\partial f / \partial x\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\ln \left(\frac{x^{2}+y}{y}\right)\)

Short Answer

Expert verified
\( \partial f / \partial x = \frac{2x}{y(x^2+y)} \), \( \partial f / \partial y = \frac{x^2}{y^2(x^2+y)} \).

Step by step solution

01

Identify the Function Components

The function is given as \( f(x, y) = \ln \left( \frac{x^{2} + y}{y} \right) \). First, identify \( u(x, y) = \frac{x^{2} + y}{y} \) which indicates that \( f(x, y) = \ln(u) \). We will use this form to differentiate \( f \) with respect to both \( x \) and \( y \).
02

Differentiate with Respect to x

Start by differentiating the inner function \( u(x, y) \) with respect to \( x \): \( \frac{\partial }{\partial x}u(x, y) = \frac{2x}{y} \). Then, use the chain rule: \( \frac{\partial f}{\partial x} = \frac{d}{du} \ln(u) \cdot \frac{\partial u}{\partial x} = \frac{1}{u} \cdot \frac{2x}{y} \). Substitute back \( u(x, y) = \frac{x^{2} + y}{y} \) to get \( \frac{\partial f}{\partial x} = \frac{2x}{y(x^2+y)} \).
03

Differentiate with Respect to y

Now, differentiate the inner function \( u(x, y) \) with respect to \( y \). Use the quotient rule: \( \frac{\partial }{\partial y}u(x, y) = \frac{y \cdot (0) - (x^2 + y) \cdot (-1)}{y^2} = \frac{x^2}{y^2} \). Again, apply the chain rule: \( \frac{\partial f}{\partial y} = \frac{1}{u} \cdot \frac{x^2}{y^2} \). Substitute back \( u(x, y) = \frac{x^{2} + y}{y} \) to get \( \frac{\partial f}{\partial y} = \frac{x^2}{y^2(x^2+y)} \).

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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 an essential tool when you're working with composite functions. You can think of a composite function as a "function within a function." For instance, if you have a function like \( f(u(x, y)) \), where \( u \) is another function, then the chain rule helps you differentiate \( f \) by taking into account how \( u \) changes when \( x \) or \( y \) change.

In simple terms, the chain rule states:
  • First, differentiate the outer function with respect to the inner function \( u \) (as if \( u \) were a standalone variable).
  • Then multiply that result by the derivative of the inner function \( u \) with respect to \( x \) or \( y \).
For our function \( f(x, y) = \ln \left( \frac{x^2 + y}{y} \right) \), we saw this in action. We first found the derivative of \( \ln(u) \), which is \( \frac{1}{u} \), and then multiplied it by the derivative of \( u(x, y) = \frac{x^2 + y}{y} \). This is how we used the chain rule to find the partial derivatives of \( f \) with respect to \( x \) and \( y \).

The chain rule is very helpful in untangling complicated functions, making it simpler to find how each variable contributes to changes in the function's output.
Quotient Rule
The quotient rule is a technique used in calculus to differentiate functions that involve division. It's a bit more complex than other differentiation rules because it involves both the numerator and the denominator. But once you get the hang of it, applying it becomes straightforward.

Here's how it works:
  • If you have a function \( u(x, y) = \frac{g(x, y)}{h(x, y)} \), then the quotient rule formula is:
  • \( \frac{d}{dx}u = \frac{g'(x, y)h(x, y) - g(x, y)h'(x, y)}{(h(x, y))^2} \)
In our exercise, we applied the quotient rule to differentiate \( u(x, y) = \frac{x^2 + y}{y} \) with respect to \( y \). The numerator, \( g(x, y) = x^2 + y \), had a derivative of \( 1 \) with respect to \( y \). The denominator, \( h(x, y) = y \), had a derivative of \( 1 \) with respect to \( y \).

After applying the quotient rule, we obtained \( \frac{x^2}{y^2} \), which we then used with the chain rule to finalize the partial derivative with respect to \( y \). The quotient rule is incredibly useful in many real-world applications when dealing with rates and changes.
Natural Logarithm Differentiation
Differentiating functions that involve natural logarithms generally requires the knowledge of a simple rule. The natural logarithm, denoted as \( \ln \), is one of the fundamental logarithmic functions typically used in calculus.When you need to differentiate \( \ln(x) \), the rule is straightforward:
  • The derivative of \( \ln(u) \) with respect to \( x \) is \( \frac{1}{u} \) multiplied by the derivative of \( u \) itself, \( u' \).
In the context of our original exercise, we deal with \( f(x, y) = \ln(\frac{x^2 + y}{y}) \). This is a composite function where \( u(x, y) = \frac{x^2 + y}{y} \).
The derivative of the natural logarithm \( \ln(u) \) is \( \frac{1}{u} \), and then applying the chain rule, you multiply by \( \frac{\partial u}{\partial x} \) or \( \frac{\partial u}{\partial y} \) as needed.

This process simplified finding the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). By breaking down exponential relationships into log functions, calculations become generally more manageable, which is why understanding how to differentiate natural logs is so valuable in calculus.

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Most popular questions from this chapter

Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(1,-1)}\left(2 x^{3}-3 y\right)(x y+1)\)

Determine the equation of the level curves \(f(x, y)=c\) and sketch the level curves for the specified values of \(c\). \(f(x, y)=\frac{x-y}{x+y} ; c=0,1,2\)

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Let $$f_{a}(x, y)=a x^{2}+y^{2}$$ for \((x, y) \in \mathbf{R}\), where \(a\) is a positive constant. (a) Assume that \(a=1\) and describe the level curves of \(f_{1}\). The graph of \(f_{1}(x, y)\) intersects both the \(x-z\) and the \(y-z\) planes; show that these two curves of intersection are parabolas. (b) Assume that \(a=4\). Then $$f_{4}(x, y)=4 x^{2}+y^{2}$$ and the level curves satisfy $$4 x^{2}+y^{2}=c$$ Use a graphing calculator to sketch the level curves for \(c=\) \(0,1,2,3\), and \(4 .\) These curves are ellipses. Find the curves of intersection of \(f_{4}(x, y)\) with the \(x-z\) and the \(y-z\) planes. (c) Repeat (b) for \(a=1 / 4\). (d) Explain in words how the surfaces of \(f_{a}(x, y)\) change when \(a\) changes.

A chemical diffuses in a container that occupies the interval \(0 \leq x \leq 1\). The concentration of the chemical at time \(t\) and at a point \(x\) is given by the diffusion equation: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} $$ (a) Suppose that the chemical is allowed to diffuse through the entire container until the concentration reaches an equilibrium value where \(c\) does not change any more with time, that is, \(\partial c / \partial t=0 .\) Suppose that chemical that touches the walls of the container is removed so that $$ c(0, t)=c(1, t)=0 . $$ The steady state concentration of chemical will be a function \(C(x)\) with $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ and \(C(0)=C(1)=0\). Show that \(C(x)=0\) satisfies this differential equation and the constraints as the points \(x=0\) and \(x=1\). (b) Now suppose that chemical is added to the container by a reaction that occurs at the wall \(x=0 .\) This reaction keeps the concentration of chemical at this wall equal to \(c(0, t)=1 . \mathrm{Un}\) der these conditions the steady state distribution of chemical will obey a differential equation: $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ with \(C(0)=1\) and \(C(1)=0 .\) Show that \(C(x)=1-x\) satisfies both the differential equation and the boundary conditions at \(x=0\) and \(x=1\). (c) Notice that the steady state distributions in (a) and (b) do not depend on \(D .\) Can you explain why?

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