Problem 30 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{4 \cdot \sqrt{\ln \sqrt{t}} \cdot t} \).

Step 1: Identify the Outer Function

The function given is \( y = \sqrt{\ln \sqrt{t}} \). Recognize it as a nested composite function. The outermost function here is the square root function applied to the expression \( \ln \sqrt{t} \). Let's express this as \( y = (\ln \sqrt{t})^{1/2} \) for differentiation purposes.

Step 2: Differentiate the Outer Function

Apply the power rule to differentiate \( y = u^{1/2} \) with respect to \( u \), where \( u = \ln \sqrt{t} \). The derivative is \( \frac{d}{du} u^{1/2} = \frac{1}{2} u^{-1/2} = \frac{1}{2}(\ln \sqrt{t})^{-1/2} \).

Step 3: Identify the Middle Function

The middle function is \( u = \ln(v) \), where \( v = \sqrt{t} \). This makes the middle function \( \ln(\sqrt{t}) \). Differentiating \( \ln(v) \) with respect to \( v \) gives \( \frac{1}{v} \).

Step 4: Differentiate the Middle Function

For \( u = \ln \sqrt{t} \), substitute back to find \( \frac{d\ln \sqrt{t}}{d\sqrt{t}} = \frac{1}{\sqrt{t}} \).

Step 5: Identify the Inner Function

The innermost function is \( v = \sqrt{t} \), which can be rewritten as \( t^{1/2} \).

Step 6: Differentiate the Inner Function

Differentiate \( v = t^{1/2} \) with respect to \( t \). Using the power rule, the derivative is \( \frac{d}{dt} t^{1/2} = \frac{1}{2}t^{-1/2} \).

Step 7: Apply the Chain Rule

Use the chain rule to combine the derivatives. Start from the outer function down to the inner: \[ \frac{dy}{dt} = \frac{1}{2}(\ln \sqrt{t})^{-1/2} \cdot \frac{1}{\sqrt{t}} \cdot \frac{1}{2}t^{-1/2}. \]

Step 8: Simplify the Expression

Simplify the expression: \[ \frac{dy}{dt} = \frac{1}{4 \cdot \sqrt{\ln \sqrt{t}} \cdot t}. \] This is the final derivative of \( y \) with respect to \( t \).

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, especially when dealing with derivative calculations involving composite functions. It comes into play when you have a function within another function and helps find derivatives in such situations. Think of the chain rule as a way to differentiate functions "layer by layer." When you encounter a composite function, you differentiate each function layer and then multiply the results. Essentially, if you have a function that is composed of functions, like \( y = f(g(x)) \), the derivative is given by:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

Here, you first differentiate the outer function \( f \), while treating the inner function \( g(x) \) as a variable, and then multiply by the derivative of the inner function. Using the chain rule might seem daunting at first, but you'll find it becomes quite intuitive with practice. It’s the go-to tool for tackling nested or layered functions in calculus problems.

Composite Function

Composite functions are essentially functions within functions. When you have more than one function acting on a variable, you form a composite function. For example, the function \( y = \sqrt{\ln \sqrt{t}} \) is a composite function because it comprises multiple layers of functions:

• The outermost function is the square root \( y = \sqrt{x} \).
• The middle function is the natural logarithm \( \ln(x) \).
• The innermost function is the square root again, applied directly to \( t \).

When working with composite functions, it’s crucial to recognize how these different functions combine and depend on one another. This recognition allows you to apply the chain rule to find derivatives. Identifying the structure of a composite function is a vital skill in calculus, helping you systematically break down a problem to find the derivative effectively.

Power Rule

The power rule is one of the most straightforward and useful differentiation rules. It applies to functions of the form \( f(x) = x^n \) where \( n \) is any real number. The power rule states that the derivative of \( x^n \) with respect to \( x \) is:

• \( \frac{d}{dx} x^n = n \cdot x^{n-1} \).

In the context of composite functions, the power rule can be nested. For instance, if you encounter \( t^{1/2} \), the power rule tells us that the derivative \( \frac{d}{dt} t^{1/2} = \frac{1}{2}t^{-1/2} \).Utilizing the power rule is quick and efficient, making it a staple in solving many calculus problems. It simplifies working with polynomial forms and is often paired with the chain rule when differentiating composite functions of more complexity.

Logarithmic Function

Logarithmic functions involve the natural logarithm, denoted as \( \ln(x) \), which serves as the inverse of the exponential function \( e^x \). The derivative of the logarithmic function \( \ln(x) \) is particularly simple:

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

In our problem, the function \( u = \ln \sqrt{t} \) is a key part of the overall composite function. Recognizing the relationship between \( u \) and the derivative \( \frac{1}{v} \) when considering \( u = \ln(v) \) helps in applying differentiation rules effectively.Understanding logarithmic differentiation is crucial, especially in problems where growth rates or multiplicative processes are involved. It often simplifies the process of differentiation when algebraic manipulation might otherwise seem cumbersome.