Problem 40 - Calculus Early Transcendentals

Question

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 \(y=e^{t}\)

Step-by-Step Solution

Answer

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 1: 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 \(t\): $$ z(t) = \ln (te^t + e^t) $$

Step 2: 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#

Step 1: 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: $$ \frac{\partial z}{\partial x} = \frac{1}{x+y},\quad \frac{\partial z}{\partial y} = \frac{1}{x+y} $$

Step 2: Differentiate x and y with respect to t

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

Step 3: 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: $$ z'(t) = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} = \frac{1}{x+y}e^t(t+1) + \frac{1}{x+y}e^t $$

Step 4: 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.

Key Concepts

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’s derivative makes solving complex expressions computationally efficient.