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Chapter 13: Problem 40

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. \(z^{\prime}(t),\) where \(z=\ln (x+y), x=t e^{t},\) and \(y=e^{t}\)

Short Answer

Expert verified
Question: Determine the derivative of the function \(z = \ln (x+y)\) with respect to \(t\), given \(x = te^t\) and \(y = e^t\). Answer: The derivative of the function with respect to \(t\) is \(z'(t) = \frac{e^t(t+1)}{te^t + e^t}\).

Step by step solution

01

Replace x and y with given functions

We are given that \(x = te^t\) and \(y = e^t\). Substitute these expressions into \(z\) to find the new expression for \(z\) in terms of \(t\): $$ z(t) = \ln (te^t + e^t) $$
02

Differentiate z with respect to t

Now we need to differentiate z with respect to \(t\). We'll do this using the derivative of the natural logarithm: $$ \frac{d}{dt}\ln(u) = \frac{u'}{u} $$ where \(u = te^t + e^t\). First, differentiate \(u\) with respect to \(t\): $$ u'(t) = \frac{d}{dt}(te^t + e^t) = e^t(t+1) $$ Now, differentiate \(z(t)\): $$ z'(t) = \frac{u'(t)}{u(t)} = \frac{e^t(t+1)}{te^t + e^t} $$ So the derivative using the substitution method is: $$ z'(t) = \frac{e^t(t+1)}{te^t + e^t} $$ #b. Chain Rule Method#
03

Differentiate z with respect to x and y

To use the Chain Rule, we first need to differentiate z with respect to x and y: $$ \frac{\partial z}{\partial x} = \frac{1}{x+y},\quad \frac{\partial z}{\partial y} = \frac{1}{x+y} $$
04

Differentiate x and y with respect to t

Next, differentiate x and y with respect to t: $$ \frac{dx}{dt} = e^t(t+1),\quad \frac{dy}{dt} = e^t $$
05

Apply the Chain Rule

Now we can apply the Chain Rule, which means to multiply the partial derivatives by the corresponding total derivatives and sum the results. We get: $$ z'(t) = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} = \frac{1}{x+y}e^t(t+1) + \frac{1}{x+y}e^t $$
06

Simplify the result

We can simplify the expression by factoring out the common terms: $$ z'(t) = \frac{e^t(t+1) + e^t}{x+y} $$ Now substitute the expressions of x and y back into the result: $$ z'(t) = \frac{e^t(t+1) + e^t}{te^t + e^t} $$ So the derivative using the Chain Rule method is: $$ z'(t) = \frac{e^t(t+1)}{te^t + e^t} $$ Both methods yield the same result for the derivative of the given function.

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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 a fundamental concept in calculus that makes it easier to differentiate composite functions. In simpler terms, a composite function is a function inside another function, like peeling an onion layer by layer. Each layer represents a function, and the chain rule helps us peel it systematically.

When using the chain rule, you first differentiate the outer function while keeping the inner function unchanged. Then, you multiply by the derivative of the inner function. This might sound complex at first, but it's about applying derivatives step-by-step.
  • Identify the outer and inner functions.
  • Differentiate the outer function, keeping the inner function as is.
  • Multiply by the derivative of the inner function.
This approach is particularly useful when dealing with functions like \( z = \ln(x+y) \), where \( x \) and \( y \) are themselves functions of another variable \( t \). By applying the chain rule, you simplify the differentiation into manageable steps.
Derivative
The derivative is at the heart of calculus and it measures how a function changes as its input changes. Imagine driving a car: the derivative is like the speedometer of your car, telling you how fast you're going (the rate of change of distance with respect to time).

Mathematically, the derivative of a function \( f(t) \) with respect to \( t \) is denoted as \( f'(t) \) or \( \frac{df}{dt} \).
  • It provides you the instantaneous rate of change.
  • Helps in finding slopes of tangent lines to curves.
  • Aids in optimizing functions and predicting future behavior.
In the given exercise, finding the derivative \( z'(t) \) involves differentiating functions like \( z = \ln(x+y) \) with respect to \( t \). This involves direct application of derivatives and methods, such as substitution or the chain rule.
Natural Logarithm
The natural logarithm, denoted as \( \ln(. ) \), is a special logarithm that is widely used in calculus and natural sciences. It has a base of \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. Because the natural logarithm is the inverse of the exponential function, it allows us to solve for time in exponential growth and decay problems.

The derivative of the natural logarithm has a unique and simple form. When differentiating \( \ln(u) \), where \( u \) is a function of \( t \), the derivative is \( \frac{1}{u} \cdot \frac{du}{dt} \).
  • Helps in simplifying the differentiation process of multiplicative functions.
  • Common in growth models and economic applications.
  • Key in solving calculus problems involving exponents.
In the context of the problem, the function \( z = \ln(x+y) \) involves finding how \( t \) affects this logarithmic relationship via differentiation. The simplicity of the logarithm鈥檚 derivative makes solving complex expressions computationally efficient.

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Most popular questions from this chapter

Use the method of your choice to ate the following limits. $$\lim _{(x, y) \rightarrow(1,0)} \frac{y \ln y}{x}$$

The angle between two planes is the angle \(\theta\) between the normal vectors of the planes, where the directions of the normal vectors are chosen so that \(0 \leq \theta<\pi\) Find the angle between the planes \(5 x+2 y-z=0\) and \(-3 x+y+2 z=0\)

Consider the following equations of quadric surfaces. a. Find the intercepts with the three coordinate axes, when they exist. b. Find the equations of the \(x y-, x z^{-}\), and \(y z\) -traces, when they exist. c. Sketch a graph of the surface. $$1-4 x^{2}+y^{2}+\frac{z^{2}}{2}=0$$

Let \(w=f(x, y, z)=2 x+3 y+4 z\), which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\), \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\).

Use the formal definition of a limit to prove that $$\lim _{(x, y) \rightarrow(a, b)}(x+y)=a+b . \text { (Hint: Take } \delta=\varepsilon / 2 \text { ) }$$
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