Problem 71 - Calculus and its applications

Question

Differentiate. $$ y=\frac{\ln x}{x^{5}} $$

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{1 - 5\ln x}{x^6} \).

Step 1: Apply the Quotient Rule

The given function is \( y = \frac{\ln x}{x^5} \). To differentiate this, you'll use the quotient rule for derivatives, which states that if you have a function \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \). Here, \( u = \ln x \) and \( v = x^5 \).

Step 2: Differentiate the Numerator and Denominator

First, differentiate \( u = \ln x \). The derivative is \( u' = \frac{1}{x} \). Next, differentiate \( v = x^5 \). The derivative is \( v' = 5x^4 \).

Step 3: Plug into the Quotient Rule Formula

Substitute \( u' = \frac{1}{x} \), \( u = \ln x \), \( v' = 5x^4 \), and \( v = x^5 \) into the quotient rule: \[ y' = \frac{\frac{1}{x} \cdot x^5 - \ln x \cdot 5x^4}{(x^5)^2} \] Simplify each part: \( u'v = x^4 \) and \( uv' = 5x^4 \ln x \).

Step 4: Simplify the Expression

Simplify the differentiated function: \[ y' = \frac{x^4 - 5x^4 \ln x}{x^{10}} \] Combine terms in the numerator: \[ y' = \frac{x^4(1 - 5\ln x)}{x^{10}} \] Further simplify by canceling \( x^4 \): \[ y' = \frac{1 - 5\ln x}{x^6} \]

Step 5: Write the Final Answer

The final derivative of the function \( y = \frac{\ln x}{x^5} \) is \[ y' = \frac{1 - 5\ln x}{x^6} \].

Key Concepts

Quotient Rule

The quotient rule is a powerful tool in calculus for differentiating a division between two functions. Imagine that when you divide one function by another, the rules for finding the derivative get a bit more complex than just differentiating each part separately. With the quotient rule, you can find the derivative of a function that is the quotient of two other functions.
For a function, say, where you have \( y = \frac{u}{v} \), the quotient rule gives you a formula to follow:

• The numerator's derivative, \( u' \), in our example is the derivative of the natural logarithm, \( \ln x \).
• The denominator's derivative, \( v' \), is simply differentiating \( x^5 \).

Then you apply the formula: \[ y' = \frac{u'v - uv'}{v^2} \] This formula helps you find how the rate of change of one function over another changes, ensuring you don't accidentally overlook any interactions.

Derivatives

Derivatives are at the heart of calculus, describing how a function changes as its input changes. To differentiate a function means to find its derivative. This gives us a mathematical way to analyze how something is changing at each point.
In our example, we are looking at the derivatives of \( \ln x \) and \( x^5 \) as parts of a larger function. Here's what happens:

• The derivative of \( \ln x \) is \( \frac{1}{x} \). This tells us how the natural log function changes.
• The derivative of \( x^5 \) is \( 5x^4 \). This indicates the rate of change of \( x^5 \) at any point.

These individual derivatives are then plugged into the quotient rule to find the overall derivative of the divided function.
By doing so, we see the derivative of the whole expression, not just the parts.

Natural Logarithm

The natural logarithm is a special type of logarithm often used in calculus and complex mathematics, represented as \( \ln x \). The base of the natural logarithm is the number \( e \), which is approximately equal to 2.718.
When you encounter \( \ln x \) in differentiation, its properties become very useful, especially since it has a straightforward derivative:

• The derivative of \( \ln x \) is \( \frac{1}{x} \).
• This derivative means that as \( x \) increases, the rate of change of \( \ln x \) decreases, making it a fascinating unique function.

In our function, \( \frac{\ln x}{x^5} \), \( \ln x \) is in the numerator. Knowing its derivative helps simplify the differentiation using the quotient rule. The simplicity of \( \ln x \)'s derivative makes it easier to plug into the quotient rule and understand how the overall function behaves.
Understanding how \( \ln x \) behaves helps untangle its behavior when used as part of more complex functions or equations.