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Magazine Chapter 7: Problem 6
In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\ln k x, k \text{ constant }$$
Short Answer
Expert verified
The derivative is \( \frac{1}{x} \).
Step by step solution
01
Apply the Chain Rule
To find the derivative of \( y = \ln(kx) \), we start by recognizing that we can apply the chain rule. The chain rule states that if \( y = \ln(u) \) where \( u \) is a function of \( x \), then \( \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \). Here, let \( u = kx \).
02
Differentiate the Inside Function
Calculate the derivative of the inside function \( u = kx \) with respect to \( x \). Since \( k \) is a constant, \( \frac{du}{dx} = k \).
03
Find the Overall Derivative
Substitute the expressions for \( \frac{1}{u} \) and \( \frac{du}{dx} \) into the chain rule formula. We have: \[ \frac{dy}{dx} = \frac{1}{kx} \cdot k = \frac{k}{kx} = \frac{1}{x} \].
04
Simplify the Expression
Simplify the result from Step 3. Since the expression \( \frac{k}{kx} \) simplifies to \( \frac{1}{x} \), this is the derivative of the function \( y = \ln(kx) \) with respect to \( x \).
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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 technique in calculus used to differentiate composite functions. It comes into play when a function is composed of two or more functions. Think of it as "unraveling" or "unwrapping" nested functions to find the rate of change. For example, when dealing with a function like \( y = \ln(kx) \), you encounter a composite structure where one function is nested inside another.
- The chain rule states that if you have a function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is given by \( f'(g(x)) \cdot g'(x) \).
- In our exercise, \( f(u) = \ln(u) \) and \( u = kx \), a new function of \( x \).
- The derivative \( f'(u) = \frac{1}{u} \) is then multiplied by the derivative of \( u \), which is \( k \), to find the overall derivative.
Logarithm
The logarithm is a mathematical concept that serves as the inverse operation to exponentiation. Natural logarithms, denoted as \( \ln \), are based on the constant \( e \approx 2.718 \). It turns multiplication into addition, which is useful for simplifying a wide range of problems.
- A logarithm log base \( b \) (of a number \( a \)) asks: "To what power must \( b \) be raised, to equal \( a \)?"
- The natural logarithm \( \ln(x) \) specifically asks, "What power do we raise \( e \) to, to obtain \( x \)?"
- This concept makes logs incredibly useful in calculus, as their unique properties simplify the differentiation process, as seen in \( y = \ln(kx) \).
Differentiation
Differentiation is a core concept in calculus that deals with finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point, akin to the concept of a slope. Let's go through a few aspects of differentiation that are essential:
- Finding the derivative involves applying rules like the power rule, product rule, quotient rule, and chain rule.
- In the specific exercise, differentiation of \( y = \ln(kx) \) with respect to \( x \) is done by recognizing its form and applying the chain rule to break it into manageable parts.
- The process leads to \( \frac{dy}{dx} = \frac{1}{x} \), indicating that for each small change in \( x \), \( y \) changes by \( \frac{1}{x} \).
Function of a Constant
A function of a constant involves expressions where constants are manipulated along with variables within a function. In calculus, the treatment of constants is straightforward yet important to grasp:
- When taking the derivative of a function, constants remain unaffected by the differentiation process unless they are in a form like \( kx \), where they multiply by a variable.
- In the exercise, \( k \) is constant, and when \( u = kx \), the derivative of \( u \) with respect to \( x \) is \( k \) because \( k \) is simply a multiplier for \( x \).
- Recognizing these constants and their roles helps simplify equations and differentiation, as seen when \( \frac{du}{dx} = k \), leading directly to a solution.
