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For any positive constant \(k,\) the derivative of \(\ln (k x)\) is 1\(/ x\) (a) by using the Chain Rule. (b) by using a property of logarithms and differentiating.

Short Answer

Expert verified
The derivative of \(\ln(kx)\) is \(\frac{1}{x}\), whether computed using the chain rule or properties of logarithms for differentiation.

Step by step solution

01

Derivative of \(\ln (k x)\) using the Chain Rule

Applying the chain rule to find the derivative of a composition of two functions, we get that the derivative of \(\ln(kx)\) where \(k\) is constant and \(x\) is the variable, is \(\frac{1}{kx} * k = \frac{1}{x}\).
02

Derivative of \(\ln (k x)\) using properties of logarithms and differentiating

By using the property of logarithms that says \(\ln(ab)=\ln(a)+\ln(b)\), we can rewrite \(\ln(kx)\) as \(\ln(k)+\ln(x)\). The derivative of \(\ln(k)\) is 0 because \(k\) is a constant, and the derivative of \(\ln(x)\) is \(\frac{1}{x}\). So, the derivative of \(\ln(kx)\) is \(\frac{1}{x}\) + 0 = \(\frac{1}{x}\).

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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 tool in calculus used for finding the derivative of a composition of functions. When you have a function nestled within another function, like in the expression \( \ln(kx) \), the chain rule comes to the rescue. It allows us to take derivatives in steps, respecting the order of the functions.

In the case of \( \ln(kx) \), consider it as \( \ln(u) \) where \( u = kx \). The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). Next, we find the derivative of \( u = kx \) with respect to \( x \), which is simply \( k \).

Now, by multiplying these two results, we apply the chain rule:
  • Derivative of \( \ln(u) = \frac{1}{u} \)
  • Derivative of \( u = kx \) is \( k \)
Putting it together, the overall derivative becomes \( \frac{1}{kx} \times k = \frac{1}{x} \).

This is a straightforward application of the chain rule, showcasing how a complicated expression can be tackled by breaking it into simpler parts.
Logarithmic Differentiation
Logarithmic differentiation is a powerful technique particularly useful when dealing with products, quotients, or powers of functions. By leveraging properties of logarithms, it transforms a potentially cumbersome problem into a manageable one.

Consider the expression \( \ln(kx) \). Through the lens of logarithmic differentiation, we can split this into:
  • \( \ln(kx) = \ln(k) + \ln(x) \)
This step employs the property of logarithms that allows the product inside the log to become a sum. Now, when you differentiate:
  • The derivative of \( \ln(k) \), a constant, is \( 0 \).
  • The derivative of \( \ln(x) \) is \( \frac{1}{x} \).
By adding these results, we find that the derivative of \( \ln(kx) \) is simply \( \frac{1}{x} \). This approach emphasizes how logarithmic rules can simplify the differentiation process, making it more accessible and less prone to errors.
Properties of Logarithms
The properties of logarithms are handy tools that simplify complex expressions, making calculus tasks more manageable. These properties transform multiplicative and divisional relationships inside logarithms into additive and subtractive ones, respectively.

Let's look at the exercise \( \ln(kx) \) and dissect it using one crucial property:
  • Product Rule: \( \ln(ab) = \ln(a) + \ln(b) \)
This property allows us to break down expressions like \( \ln(kx) \) into \( \ln(k) + \ln(x) \).

Another important characteristic is:
  • Constant Rule: The logarithm of a constant, like \( \ln(k) \), does not change when differentiated, resulting in \( 0 \).
Understanding these properties not only helps in differentiation but also in solving equations and simplifying calculations throughout calculus. They form a foundational pillar that supports various advanced mathematical techniques, making them indispensable in a student's toolkit.

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