Question

Assume \(y>0\) is fixed and \(x>0 .\) Show that \(\frac{d}{d x}(\ln x y)=\frac{d}{d x}(\ln x) .\) Recall that if two functions have the same derivative, then they differ by an additive constant. Set \(x=1\) to evaluate the constant and prove that \(\ln x y=\ln x+\ln y\).

Short answer

A: The result of differentiating ln(xy) with respect to x is 1/x.

Step-by-step solution

Differentiate \(\ln(xy)\) with respect to \(x\)

Using the chain rule, we have: \[\frac{d}{dx}(\ln(xy)) = \frac{1}{xy}\cdot\frac{d}{dx}(xy)\]

Calculate \(\frac{d}{dx}(xy)\)

Since \(y\) is a constant and \(x\) is the variable, we get: \[\frac{d}{dx}(xy) = y\cdot\frac{d}{dx}(x) = y\]

Substitute the value of \(\frac{d}{dx}(xy)\) back into Step 1

Now, use the value found in Step 2 and substitute it back into the first equation: \[\frac{d}{dx}(\ln(xy)) = \frac{1}{xy}\cdot y\]

Simplify the expression

The expression simplifies to: \[\frac{d}{dx}(\ln(xy)) = \frac{1}{x}\]

Differentiate \(\ln(x)\) with respect to \(x\)

\[\frac{d}{dx}(\ln(x)) = \frac{1}{x}\]

Compare the derivatives

We can see that: \[\frac{d}{dx}(\ln(xy)) = \frac{d}{dx}(\ln(x))\]

Set \(x = 1\)

Now we need to show that \(\ln(xy) = \ln(x) + \ln(y)\). Since the derivatives of both functions are equal, they differ by an additive constant. Set \(x = 1\) and evaluate the constant: \[\ln(1y) = \ln(1) + C\] \[\ln(y) = C\]

Prove \(\ln(xy) = \ln(x) + \ln(y)\)

So, since the constant is equal to \(\ln(y)\), we have: \[\ln(xy) = \ln(x) + \ln(y)\] We have shown that the derivative of \(\ln(xy)\) is equal to the derivative of \(\ln(x)\), and that the functions only differ by an additive constant (which is \(\ln(y)\)). Therefore, we have proven that \(\ln(xy) = \ln(x) + \ln(y)\).

Key concepts

Chain Rule

The chain rule is a powerful tool in calculus that helps us differentiate complex functions. When we have a composite function, the chain rule allows us to differentiate by focusing on the layers separately.
In the case of finding the derivative of \( \ln(xy) \), the chain rule is applied by first acknowledging \( xy \) as a product inside the logarithmic function. This means we need to differentiate the outer function, which is the natural logarithm, and then multiply by the derivative of the inner function, \( xy \).
Using the formula for the derivative of a natural logarithm, \( \frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx} \), for \( u = xy \), we find:

• First, differentiate \( \ln(xy) \) as \( \frac{1}{xy} \cdot \frac{d}{dx}(xy) \).
• Second, calculate \( \frac{d}{dx}(xy) \) where \( y \) is constant, resulting in simply \( y \).
• Finally, plug back \( y \) to get the simplified result: \( \frac{1}{x} \).

By using the chain rule, we break down the process into manageable parts, streamlining complex differentiations.

Derivative of Logarithm

Understanding the derivative of a logarithmic function is key to mastering calculus. The natural logarithm, denoted as \( \ln(x) \), simplifies the process of differentiation due to its straightforward derivative.
The derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \). This formula arises because the natural logarithm is the inverse of the exponential function. Its unique properties allow differentiation to produce simple reciprocals.
When we differentiate a combined function like \( \ln(xy) \), we apply this basic rule of the natural logarithm by first recognizing the inside function, adjusting our differentiation procedure to include all factors of \( x \) and \( y \).
In our exercise, we found:

• The derivative of \( \ln(xy) \) simplifies to \( \frac{1}{x} \), establishing it is identical to the derivative of \( \ln(x) \).
• By showing \( \frac{d}{dx}(\ln(xy)) = \frac{d}{dx}(\ln(x)) \), we confirm the equality based on their derivatives.

This underlying simplicity in the derivative of logarithms makes them easier to use in calculus, particularly with multiplying and dividing functions.

Properties of Logarithms

Logarithms come with distinct properties that often simplify calculations and transformations of equations. Recognizing and applying these properties is crucial in many mathematical contexts, especially in calculus.
For this exercise, an important property is the product property of logarithms, expressed as:\[\ln(xy) = \ln(x) + \ln(y)\]This property shows how multiplication inside a logarithm can be separated into addition outside of the logarithm, facilitating calculations in equations involving products.
The solution verifies this by setting \( x = 1 \). This specific setup reveals a constant \( C \) equal to \( \ln(y) \), establishing the relationship: \[\ln(xy) = \ln(x) + \ln(y)\]
With this approach, logarithmic properties enable us to simplify complex functions into more manageable components, useful for proving identity and equivalence in calculus. Logarithm properties like power, quotient, and change of base are similarly useful across different mathematical challenges.