Problem 26 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\frac{t}{\sqrt{\ln t}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{2(\ln t) - 1}{2(\ln t)\sqrt{\ln t}} \).

Step 1: Identify the function and the rule to use

The given function is \( y = \frac{t}{\sqrt{\ln t}} \). This can be seen as a quotient, which suggests using the quotient rule for differentiation: if \( u(t) = t \) and \( v(t) = \sqrt{\ln t} \), then \( y = \frac{u}{v} \).

Step 2: Differentiate the numerator and denominator

Differentiate the numerator: \( u(t) = t \), thus \( u'(t) = 1 \).Differentiate the denominator: \( v(t) = \sqrt{\ln t} = (\ln t)^{1/2} \). Using the chain rule, the derivative \( v'(t) = \frac{1}{2}(\ln t)^{-1/2} \cdot \frac{1}{t} \).

Step 3: Apply the Quotient Rule

The quotient rule states that the derivative of \( \frac{u}{v} \) is \( \frac{u'v - uv'}{v^2} \).Substitute the derivatives and the original functions:\[\frac{d}{dt} \left( \frac{t}{\sqrt{\ln t}} \right) = \frac{1 \cdot \sqrt{\ln t} - t \cdot \frac{1}{2}(\ln t)^{-1/2} \cdot \frac{1}{t}}{(\sqrt{\ln t})^2}\].

Step 4: Simplify the Expression

Simplify the numerator and the denominator:The numerator becomes \( \sqrt{\ln t} - \frac{1}{2\sqrt{\ln t}} \).The denominator becomes \( \ln t \).Thus, the derivative is:\[\frac{\sqrt{\ln t} - \frac{1}{2\sqrt{\ln t}}}{\ln t} \].

Step 5: Combine the Terms

Express the terms in the numerator over a common denominator:\[= \frac{2(\ln t) - 1}{2(\ln t)\sqrt{\ln t}}\].

Key Concepts

Derivative

A derivative in calculus measures how a function changes as its input changes. It's like finding the slope of a curve at any given point.
In this exercise, we are tasked with finding the derivative of the equation \( y = \frac{t}{\sqrt{\ln t}} \). This means we're looking at how \( y \) changes as \( t \) changes.
Understanding derivatives is crucial as they tell us if a function is increasing or decreasing, and by how much.

• The derivative of a linear function is its slope, which is a constant.
• For more complex functions like the one in this exercise, we use rules like the Quotient and Chain Rules to determine the derivative.

Quotient Rule

The Quotient Rule is an essential tool in calculus, which we use when we need to differentiate a function where one function is divided by another.
For this rule, if \( u(t) \) and \( v(t) \) are two functions, then \( \frac{u}{v} \) is their quotient. The rule states:
\[\frac{d}{dt}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\]
This exercise demonstrates the use of the Quotient Rule with:

• \( u(t) = t \) with its derivative \( u'(t) = 1 \)
• \( v(t) = \sqrt{\ln t} \) further differentiated using the Chain Rule.

Applying the Quotient Rule helps us systematically determine the derivative of complex fractions.

Chain Rule

The Chain Rule in calculus is used to differentiate composite functions, that is, functions within other functions.
This rule comes into play when we differentiated \( v(t) = \sqrt{\ln t} \). We rewrite it as \( (\ln t)^{1/2} \). The Chain Rule gives us a way to find the derivative by multiplying the derivative of the outer function by the derivative of the inner function.
Here, by recognizing \( \ln t \) as the inner function, the derivative becomes:
\[v'(t) = \frac{1}{2}(\ln t)^{-1/2} \cdot \frac{1}{t}\]
This step is crucial for accurately applying the Quotient Rule later.

Simplification

Simplification in the context of calculus involves making derived expressions more straightforward.
Once the Quotient Rule has been applied, the derivative often appears cumbersome and complex. To communicate the solution effectively and solve problems easily, simplification is necessary.
In this problem, the expression after applying the Quotient Rule is:
\[\frac{\sqrt{\ln t} - \frac{1}{2\sqrt{\ln t}}}{\ln t}\]
We then further express and simplify it by finding a common denominator:
\[= \frac{2(\ln t) - 1}{2(\ln t)\sqrt{\ln t}}\]

• This makes the expression cleaner and easier to work with.
• Benefits include easier interpretation and manipulation of the derivative in further calculations.

Logarithms

Logarithms are mathematical operations that are essentially the inverse of exponents. In the given problem, the logarithmic function \( \ln t \) is part of the expression we are differentiating.
Understanding how to differentiate \( \ln t \) is key to solving this problem, especially when it is nested within other functions as seen with \( \sqrt{\ln t} \).
Key points about differentiating logarithms include:

• \( \frac{d}{dt} \ln t = \frac{1}{t} \) is the basic differentiation formula for natural logarithms.
• Logarithms can convert multiplication into addition, which can simplify calculations.

In this exercise, knowing the properties of logarithms allows integrating them into the Chain Rule application effectively.