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Chapter 14: Problem 16

In Exercises \(1-22,\) find \(\partial f / \partial x\) and \(\partial f / \partial y\) $$f(x, y)=e^{x y} \ln y$$

Short Answer

Expert verified
\(\frac{\partial f}{\partial x} = y e^{x y} \ln y\) and \(\frac{\partial f}{\partial y} = e^{x y} \left( x \ln y + \frac{1}{y} \right)\).

Step by step solution

01

Identify the function components

The given function is \(f(x, y) = e^{x y} \ln y\). This function involves two parts: an exponential function \(e^{x y}\) and a logarithmic function \(\ln y\). Both are a function of two variables, \(x\) and \(y\).
02

Find \(\partial f/\partial x\)

To find the partial derivative of \(f\) with respect to \(x\), treat \(y\) as a constant. Use the product rule, since \(f(x, y)\) consists of the product of \(e^{x y}\) and \(\ln y\):\[\frac{\partial f}{\partial x} = \left(\frac{\partial}{\partial x} e^{x y}\right) \ln y + e^{x y} \left(\frac{\partial}{\partial x} \ln y\right)\]Compute:- \(\frac{\partial}{\partial x} e^{x y} = y e^{x y}\) (since the derivative of \(x y\) with respect to \(x\) is \(y\))- \(\frac{\partial}{\partial x} \ln y = 0\) (since \(\ln y\) is a constant with respect to \(x\))Thus,\[\frac{\partial f}{\partial x} = y e^{x y} \ln y\]
03

Find \(\partial f/\partial y\)

To find the partial derivative of \(f\) with respect to \(y\), treat \(x\) as a constant. Use the product rule again:\[\frac{\partial f}{\partial y} = \left(\frac{\partial}{\partial y} e^{x y}\right) \ln y + e^{x y} \left(\frac{\partial}{\partial y} \ln y\right)\]Compute:- \(\frac{\partial}{\partial y} e^{x y} = x e^{x y}\) (since the derivative of \(x y\) with respect to \(y\) is \(x\))- \(\frac{\partial}{\partial y} \ln y = \frac{1}{y}\)Thus,\[\frac{\partial f}{\partial y} = x e^{x y} \ln y + e^{x y} \frac{1}{y} = e^{x y} \left( x \ln y + \frac{1}{y} \right)\]

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Key Concepts

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

Product Rule
When you have a function that is the product of two or more functions, you use the product rule to find its derivative. This rule is like a recipe that tells us how to differentiate a product of functions.
This is especially useful in functions composed of terms like \[ f(x, y) = e^{xy} \ln y \] where two parts are multiplying. Let's break it down:
  • Suppose your function is \( u(x, y) \cdot v(x, y) \).
  • To find the derivative with respect to a variable (like \(x\) or \(y\)), apply the product rule: \[ \frac{\partial}{\partial x}(uv) = u'v + uv' \]
  • For \( \partial f/\partial x \), if \(u\) is \(e^{xy}\) and \(v\) is \(\ln y\), differentiate each part separately while treating the other one as a constant.
This approach helps us find each part's role in the whole function, ensuring we capture changes accurately.
Remember, familiarity with products allows you to tackle complexities in math one step at a time, making it easier to understand and solve.
Exponential Function
The exponential function, especially when involving powers of \(e\), is a cornerstone in math, widely used to model growth processes. In our equation, the term \[ e^{xy} \] stands out as an exponential function. Here鈥檚 why it matters:
  • The base \(e\) is a unique number, approximately 2.718, that arises naturally in various processes.
  • This function can dramatically change the values of the output even with small changes in \(x\) and \(y\).
  • When differentiating \( e^{xy} \) with respect to \(x\), remember: the derivative of \( e^{u} \) with respect to \(x\) is \( u'e^{u} \).
For our purposes, \( u = xy \), so differentiating with respect to \(x\) gives \( ye^{xy} \), observing how the change in \(x\) affects the function.
The exponential's property of maintaining its form even after differentiation simplifies operations, especially in calculus.
Logarithmic Function
Logarithmic functions like \( \ln y \) in our example are incredibly useful, converting multiplication into addition. This property simplifies many calculations, especially when dealing with exponential growth.
In calculus, the derivative of a logarithmic function provides crucial information:
  • When we differentiate \( \ln y \) with respect to \(y\), we get \( \frac{1}{y} \).
  • This derivative shows how a small increase in \( y \) changes the function's output.
  • Notice that \( \ln y \) doesn't change with respect to \(x\), its derivative in \( \partial f/\partial x \) is zero.
This neutrality towards changes in \(x\) simplifies our task while differentiating complex functions involving logarithms.
By understanding these elements, the role of the logarithmic function becomes clear in various scenarios, including solving partial derivatives efficiently.

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Most popular questions from this chapter

Let $$f(x, y)=\left\\{\begin{array}{ll}{\frac{x y^{2}}{x^{2}+y^{4}},} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.$$ Show that \(f_{x}(0,0)\) and \(f_{y}(0,0)\) exist, but \(f\) is not differentiable at \((0,0) .\) (Hint: Use Theorem 4 and show that \(f\) is not continuous at \((0,0) . )\)

Does a function \(f(x, y)\) with continuous first partial derivatives throughout an open region \(R\) have to be continuous on \(R ?\) Give reasons for your answer.

Extrema on a curve of intersection Find the extreme values of \(f(x, y, z)=x^{2} y z+1\) on the intersection of the plane \(z=1\) with the sphere \(x^{2}+y^{2}+z^{2}=10\)

If \(z=x+f(u),\) where \(u=x y,\) show that \begin{equation}x \frac{\partial z}{\partial x}-y \frac{\partial z}{\partial y}=x.\end{equation}

Suppose that \(f(x, y, z, w)=0\) and \(g(x, y, z, w)=0\) determine \(z\) and \(w\) as differentiable functions of the independent variables \(x\) and \(y,\) and suppose that \begin{equation} \frac{\partial f}{\partial z} \frac{\partial g}{\partial w}-\frac{\partial f}{\partial w} \frac{\partial g}{\partial z} \neq 0.\end{equation} Show that \begin{equation}\left(\frac{\partial z}{\partial x}\right)_{y}=-\frac{\frac{\partial f}{\partial x} \frac{\partial g}{\partial w}-\frac{\partial f}{\partial w} \frac{\partial g}{\partial x}}{\frac{\partial f}{\partial z} \frac{\partial g}{\partial w}-\frac{\partial f}{\partial w} \frac{\partial g}{\partial z}}\end{equation} and \begin{equation} \left(\frac{\partial w}{\partial y}\right)_{x}=-\frac{\frac{\partial f}{\partial z} \frac{\partial g}{\partial y}-\frac{\partial f}{\partial y} \frac{\partial g}{\partial z}}{\frac{\partial f}{\partial z} \frac{\partial g}{\partial w}-\frac{\partial f}{\partial w} \frac{\partial g}{\partial z}}. \end{equation}
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