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Chapter 12: Problem 40

Derivative practice two ways Find the indicated derivative in two ways: a. Replace \(x\) and \(y\) to write \(z\) as a function of t and differentiate. b. Use the Chain Rule. $$z^{\prime}(t), \text { where } z=\ln (x+y), x=t e^{t}, \text { and } y=e^{t}$$

Short Answer

Expert verified
Question: Find the derivative of the function \(z = \ln(x+y)\) with respect to variable \(t\), where \(x = te^t\) and \(y = e^t\), using two methods: direct differentiation and the Chain Rule. Answer: The derivative of \(z\) with respect to \(t\) for both methods is \(z'(t) = 1 + \frac{e^t}{te^t + e^t}\).

Step by step solution

01

Way 1: Direct Differentiation

First, we need to replace \(x\) and \(y\) in the equation \(z = \ln(x+y)\) by their expressions in terms of \(t\). Then we will directly differentiate \(z(t)\) with respect to \(t\): \(x = te^t\) \(y = e^t\) \(z = \ln(te^t + e^t)\) Now, differentiate \(z\) with respect to \(t\): \(z'(t) = \frac{1}{te^t + e^t} \cdot \frac{d(te^t + e^t)}{dt}\) To find the derivate of \((te^t + e^t)\) with respect to \(t\), we can use some rules of differentiation for sum and product of functions: \(\frac{d(te^t + e^t)}{dt} = t\frac{d(e^t)}{dt} + e^t\frac{dt}{dt} + \frac{d(t)}{dt}e^t + \frac{d(e^t)}{dt}\) Since \(\frac{d(e^t)}{dt}=e^t\), \(\frac{dt}{dt}=1\), and \(\frac{d(t)}{dt}=1\), the equation above becomes: \(\frac{d(te^t + e^t)}{dt} = te^t + e^t + e^t\) Now substitute this back into \(z'(t)\): \(z'(t) = \frac{1}{te^t + e^t} \cdot (te^t + e^t + e^t) = \frac{te^t + e^t + e^t}{te^t + e^t} = \boxed{1 + \frac{e^t}{te^t + e^t}}\)
02

Way 2: Chain Rule

Now, we will use the Chain Rule to find the derivative \(z'(t)\). The chain rule states that if \(z\) is a composite function of \(x\) and \(y\), then: \(z'(t) = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}\) First, calculate the partial derivatives: \(\frac{\partial z}{\partial x} = \frac{1}{x+y}\) \(\frac{\partial z}{\partial y} = \frac{1}{x+y}\) Then, calculate the derivatives of \(x\) and \(y\): \(\frac{dx}{dt} = e^t + te^t\) \(\frac{dy}{dt} = e^t\) Now substitute these back into the Chain Rule formula: \(z'(t) = \frac{1}{x+y}(e^t + te^t) + \frac{1}{x+y}(e^t)\) Since \(x+y=te^t+e^t\), we can write: \(z'(t) = \frac{1}{te^t+e^t}(e^t + te^t) + \frac{1}{te^t+e^t}(e^t) = \boxed{1 + \frac{e^t}{te^t + e^t}}\) Both ways give us the same derivative for \(z'\): \(z'(t) = 1 + \frac{e^t}{te^t + e^t}\)

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Key Concepts

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

Derivative
Derivatives are a fundamental concept in calculus. They represent the rate at which a function is changing at any given point. If you think of a graph, the derivative at a point is the slope of the tangent line there. In the given exercise, we use derivatives to determine how the function \( z = \ln(x+y) \) changes as the variable \( t \) changes.
To calculate a derivative like \( z'(t) \), you follow a process of differentiation. In this exercise's first solution method, we replace \( x \) and \( y \) with expressions in terms of \( t \) and then directly differentiate \( z \, ext{w.r.t} \, t \).

Here's a brief recap:
  • Find the expression for \( z \) with the desired variable, here \( t \).
  • Compute \( \frac{dz}{dt} \) using rules of differentiation like the sum rule and product rule.
  • The final expression tells you how \( z \) changes concerning \( t \).

Understanding derivatives is crucial as they are the building blocks for discovering rates and solving problems that involve both linear and non-linear changes.
Partial Derivatives
Partial derivatives apply when you have functions with several variables, like \( z = \ln(x+y) \). Instead of determining the rate of change with respect to one variable, partial derivatives look at the rate of change while keeping others constant. In our problem, \( z \) is dependent on both \( x \) and \( y \), which in turn depend on \( t \).
We use symbols like \( \frac{\partial z}{\partial x} \) to show partial derivatives. This notation signifies the derivative of \( z \) with respect to \( x \), holding \( y \) constant. For \( z \) in our exercise, you calculate:
  • A partial derivative \( \frac{\partial z}{\partial x} = \frac{1}{x+y} \) which looks at the change when altering \( x \) only.
  • Similar calculation for \( \frac{\partial z}{\partial y} = \frac{1}{x+y} \).
Combining partial derivatives with respect to individual variables provides a comprehensive view of how a function behaves as each variable changes.
Partials are especially useful in multivariable functions, enabling more structured solutions in real-world applications like physics and engineering.
Function Composition
Function composition involves creating a complex function by combining simpler functions. In mathematical terms, if you have two functions \( f(x) \) and \( g(x) \), their composition is denoted as \((f \circ g)(x) = f(g(x))\). The Chain Rule, which we use in this exercise, is essentially about differentiating such composite functions.
For \( z = \ln(x+y) \), think of \( x \) and \( y \) as individual functions of \( t \). You are essentially nesting them inside \( z \).
To apply the Chain Rule:
  • Determine the partial derivatives of \( z \) with respect to \( x \) and \( y \).
  • Find derivatives of \( x \) and \( y \) with respect to \( t \).
  • Plug these into the chain rule formula to get \( z'(t) \).
The result is the rate at which \( z \) changes concerning \( t \) when \( x \) and \( y \) depend on \( t \).
Understanding composition allows you to fluidly move between functions and carefully unpack how each impacts the others, being key in scenarios where different related variables define a single outcome.

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Most popular questions from this chapter

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y)=x y \cos x y$$

Let \(h\) be continuous for all real numbers. a. Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=\int_{x}^{y} h(s) d s\). b. Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=\int_{1}^{x y} h(s) d s\).

(Adapted from 1938 Putnam Exam) Consider the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) and the plane \(P\) given by \(A x+B y+C z+1=0\). Let \(h=\left(A^{2}+B^{2}+C^{2}\right)^{-1 / 2}\) and \(m=\left(a^{2} A^{2}+b^{2} B^{2}+c^{2} C^{2}\right)^{1 / 2}\) a. Find the equation of the plane tangent to the ellipsoid at the point \((p, q, r)\) b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\) and find equations of the tangent planes. c. Show that the distance between the origin and the plane \(P\) is \(h\) d. Show that the distance between the origin and the tangent planes is \(h m\) e. Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid.

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Let the equation of the best-fit line be \(y=m x+b,\) where the slope \(m\) and the \(y\) -intercept \(b\) must be determined using the least squares condition. First assume that there are three data points \((1,2),(3,5),\) and \((4,6) .\) Show that the function of \(m\) and \(b\) that gives the sum of the squares of the vertical distances between the line and the three data points is $$ \begin{aligned} E(m, b)=&((m+b)-2)^{2}+((3 m+b)-5)^{2} \\ &+((4 m+b)-6)^{2} \end{aligned}. $$ Find the critical points of \(E\) and find the values of \(m\) and \(b\) that minimize \(E\). Graph the three data points and the best-fit line.

a. Show that the point in the plane \(a x+b y+c z=d\) nearest the origin is \(P\left(a d / D^{2}, b d / D^{2}, c d / D^{2}\right),\) where \(D^{2}=a^{2}+b^{2}+c^{2} .\) Conclude that the least distance from the plane to the origin is \(|d| / D\). (Hint: The least distance is along a normal to the plane.) b. Show that the least distance from the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) to the plane \(a x+b y+c z=d\) is \(\left|a x_{0}+b y_{0}+c z_{0}-d\right| / D\) (Hint: Find the point \(P\) on the plane closest to \(P_{0}\).)
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