/*! 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 12 Find the gradient of the functio... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the gradient of the function. $$f(x, y, z)=e^{z^{2}}+y \ln \left(x^{2}+5\right)$$

Short Answer

Expert verified
The gradient is \( \left( \frac{2xy}{x^2+5}, \ln(x^2+5), 2z e^{z^2} \right) \).

Step by step solution

01

Understand the Function

Given the function \( f(x, y, z) = e^{z^2} + y \ln(x^2 + 5) \), it is a multivariable function composed of exponential and logarithmic terms. To find the gradient, we need to compute the partial derivatives with respect to each variable \( x \), \( y \), and \( z \).
02

Find the Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) and \( z \) as constants. The derivative of \( y \ln(x^2 + 5) \) with respect to \( x \) is \( y \cdot \frac{1}{x^2 + 5} \cdot 2x \) which simplifies to \( \frac{2xy}{x^2 + 5} \). Since \( e^{z^2} \) doesn't involve \( x \), its derivative is 0.
03

Find the Partial Derivative with Respect to y

For \( \frac{\partial f}{\partial y} \), treat \( x \) and \( z \) as constants. The partial derivative is just the constant multiplier, which is \( \ln(x^2 + 5) \). Since \( e^{z^2} \) is independent of \( y \), its derivative is 0.
04

Find the Partial Derivative with Respect to z

To find \( \frac{\partial f}{\partial z} \), consider \( x \) and \( y \) as constants. The derivative of \( e^{z^2} \) with respect to \( z \) is obtained using the chain rule: \( 2z e^{z^2} \). The \( y \ln(x^2 + 5) \) term does not depend on \( z \), so its derivative is 0.
05

Express the Gradient

Now combine these results to express the gradient: \( abla f(x, y, z) = \left( \frac{2xy}{x^2 + 5}, \ln(x^2 + 5), 2z e^{z^2} \right) \).

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Key Concepts

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

Partial Derivatives
Partial derivatives are a cornerstone concept in multivariable calculus. They help us understand how a function changes as one of its variables changes, while keeping the other variables constant. When we have a function of several variables, like our function, \( f(x, y, z) = e^{z^2} + y \ln(x^2 + 5) \), finding the derivative with respect to one variable gives us the rate of change in that direction.
  • To find the partial derivative with respect to \( x \), you treat \( y \) and \( z \) as constants. This means you only consider how changes in \( x \) affect the function.
  • Similarly, to find the partial derivative with respect to \( y \), treat \( x \) and \( z \) as constants.
  • Lastly, for the partial derivative with respect to \( z \), \( x \) and \( y \) are treated as constants.
Understanding this helps in visualizing the gradient of the function, acting as an arrow pointing in the direction of the greatest increase in function value.
Chain Rule
The chain rule is a powerful tool for differentiation, especially when dealing with composite functions like our exponential term, \( e^{z^2} \). The basic idea behind the chain rule is that you can break down the differentiation of a composite function into simpler parts, which are easier to manage.
When applying the chain rule to \( e^{z^2} \) with respect to \( z \), we deal with the outer function \( e^{u} \), where \( u = z^2 \), and the inner function \( z^2 \).
  • The derivative of the outer function \( e^{u} \) with respect to \( u \) is \( e^u \).
  • The derivative of the inner function \( z^2 \) with respect to \( z \) is \( 2z \).
  • By multiplying these, we get \( 2z e^{z^2} \) as the partial derivative of \( e^{z^2} \) with respect to \( z \).
The chain rule breaks down the problem into manageable steps, showing its versatility in calculus.
Exponential Functions
Exponential functions are everywhere in mathematical modeling, defined generally as \( e^x \). They have distinct properties, most notably the fact that the function and its derivative are the same in form. For example, \( \frac{d}{dx} e^{x} = e^{x} \).
In our function \( f(x, y, z) = e^{z^2} + y \ln(x^2 + 5) \), the exponential function \( e^{z^2} \) is crucial in the partial derivative with respect to \( z \). Using the chain rule, we find its derivative as \( 2z e^{z^2} \), demonstrating how exponentials carry their own shape through differentiation.
Exponential functions grow quickly. They effectively represent growth processes like population increase or compound interest in economics. They also often involve the mathematical constant \( e \), approximately 2.718, which naturally emerges in various growth problems.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions. In essence, the natural logarithm \( \ln(x) \) answers the question: "To what power must \( e \) be raised to get \( x \)?". A significant property is its derivative, \( \frac{d}{dx} \ln(x) = \frac{1}{x} \), reflecting the rate of change of logarithmic growth.
In our given function \( f(x, y, z) = e^{z^2} + y \ln(x^2 + 5) \), the term \( y \ln(x^2 + 5) \) involves a logarithm. When we differentiate it with respect to \( x \), the derivative term \( \frac{2xy}{x^2 + 5} \) manifests, using the chain rule to accommodate the \( x^2+5 \) argument of the logarithm.
Logarithmic functions often model phenomena like sound intensity or acidity in chemistry, as they can represent saturation or decay processes effectively. By understanding \( y \ln(x^2 + 5) \), you're grasping how logarithms help solve real-world problems by translating them into a mathematical form that's easier to manipulate and comprehend.

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Most popular questions from this chapter

Are the statements true or false? Give measons for your answer. The tangent plane approximation of \(f(x, y)=y e^{x^{2}}\) at the point (0,1) is \(f(x, y) \approx y\)

Find the gradient of the function. Assume the variables are restricted to a domain on which the function s defined. $$f(x, y)=\ln \left(x^{2}+y^{2}\right)$$

The tastiness, \(T\), of a soup depends on the volume, \(V\) of the soup in the pot and the quantity, \(S,\) of salt in the soup. If you have more soup, you need more salt to make it taste good. Match the three stories (a)-(c) to the three statements (I)-(III) about partial derivatives. (a) I started adding salt to the soup in the pot. At first the taste improved, but eventually the soup became too salty and continuing to add more salt made it worse. (b) The soup was too salty, so I started adding unsalted soup. This improved the taste at first, but eventually there was too much soup for the salt, and continuing to add unsalted soup just made it worse. (c) The soup was too salty, so adding more salt would have made it taste worse. I added a quart of unsalted soup instead. Now it is not salty enough, but I can improve the taste by adding salt. (I) \(\partial^{2} T / \partial V^{2} < 0\) (II) \(\partial^{2} T / \partial S^{2} < 0\) (III) \(\partial^{2} T / \partial V \partial S > 0\)

Find the gradient of the function. Assume the variables are restricted to a domain on which the function s defined. $$f(m, n)=m^{2}+n^{2}$$

A student was asked to find the directional derivative of \(f(x, y)=x^{2} e^{y}\) at the point (1,0) in the direction of \(\vec{v}=4 \vec{i}+3 \vec{j} .\) The student's answer was $$ f_{i i}(1,0)=\operatorname{grad} f(1,0) \cdot \vec{u}=\frac{8}{5} \vec{i}+\frac{3}{5} \vec{j} $$ (a) At a glance, how do you know this is wrong? (b) What is the correct answer?

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