Problem 16 - Thomas Calculus

Question

In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=t \sqrt{\ln t} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dt} = \frac{2\ln t + 1}{2\sqrt{\ln t}} \).

Step 1: Identify the function type

The given function is written as a product: \( y = t \sqrt{\ln t} \). This can be considered a product of two separate functions, \( f(t) = t \) and \( g(t) = \sqrt{\ln t} \). We will use the product rule to differentiate this function.

Step 2: Recall the product rule

The product rule states that if \( y = u(t) \cdot v(t) \), then the derivative \( \frac{dy}{dt} \) is given by \( u'(t)v(t) + u(t)v'(t) \). Set \( u(t) = t \) and \( v(t) = \sqrt{\ln t} \). We will need both \( u'(t) \) and \( v'(t) \).

Step 3: Differentiate \( u(t) \)

The function \( u(t) = t \) is a basic polynomial. Thus, its derivative with respect to \( t \) is \( u'(t) = 1 \).

Step 4: Differentiate \( v(t) \)

We have \( v(t) = (\ln t)^{1/2} \). Use the chain rule to differentiate it: \( v'(t) = \frac{1}{2}(\ln t)^{-1/2} \cdot \frac{1}{t} = \frac{1}{2t\sqrt{\ln t}} \).

Step 5: Apply the product rule

Substitute \( u(t), u'(t), v(t), \) and \( v'(t) \) into the product rule: \( \frac{dy}{dt} = 1 \cdot \sqrt{\ln t} + t \cdot \frac{1}{2t\sqrt{\ln t}} \).

Step 6: Simplify the expression

Simplify the expression: \( \frac{dy}{dt} = \sqrt{\ln t} + \frac{1}{2\sqrt{\ln t}} \). Combine these under a common denominator: \( \frac{dy}{dt} = \frac{2\ln t + 1}{2\sqrt{\ln t}} \).

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus that helps us find the derivative of a product of two functions. It is particularly useful when a function is a multiplication of two separate functions, like in the problem where we have the expression \( y = t \sqrt{\ln t} \). Here, we can recognize this as the product of \( f(t) = t \) and \( g(t) = \sqrt{\ln t} \).

To apply the product rule, it is crucial to recall its form: if \( y = u(t) \cdot v(t) \), then the derivative \( \frac{dy}{dt} \) is given by:

• \( u'(t)v(t) + u(t)v'(t) \)

This means we differentiate each function separately:
- \( u(t) \) becomes \( u'(t) \)- \( v(t) \) becomes \( v'(t) \)

After computing these derivatives, the product rule combines them. This approach simplifies the process of differentiating complex expressions.

Chain Rule

The chain rule is another essential tool in differentiation, especially when dealing with composite functions. When a function is nested inside another function, like \( v(t) = \sqrt{\ln t} \), the chain rule allows us to break the problem into simpler parts.

For the expression \( \sqrt{\ln t} \), we see it as two functions: the square root function applied to the natural logarithm function. Here's how the chain rule works for \( v(t) = (\ln t)^{1/2} \):

• Differentiate the outer function: The derivative of \( x^{1/2} \) is \( \frac{1}{2}x^{-1/2} \)
• Differentiate the inner function: The derivative of \( \ln t \) is \( \frac{1}{t} \)
• Multiply these derivatives: The chain rule gives \( v'(t) = \frac{1}{2}(\ln t)^{-1/2} \cdot \frac{1}{t} \)

This results in \( v'(t) = \frac{1}{2t\sqrt{\ln t}} \), showing how the chain rule elegantly handles layered functions.

Logarithmic Functions

Logarithmic functions appear frequently in calculus and come with their own rules for differentiation. The function \( \ln t \), or the natural logarithm of \( t \), is a common example. In finding the derivative of such functions, there's a simple rule:

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

This rule states that the derivative of the natural logarithm of a variable is the reciprocal of that variable.

When logarithmic functions are part of a more complex function, like in our problem with \( \sqrt{\ln t} \), they are often coupled with other differentiation techniques like the chain rule or product rule, as we've seen. Logarithmic functions simplify mathematical modeling by turning multiplication into addition and can be invaluable in calculus for easing the differentiation and integration processes.