Problem 14 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln \left(t^{3 / 2}\right)+\sqrt{t}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = \frac{3}{2t} + \frac{1}{2\sqrt{t}} \)

Step 1: Break Down the Function

We are given the function \( y = \ln(t^{3/2}) + \sqrt{t} \). This function is composed of two distinct parts: 1. \( \ln(t^{3/2}) \) - a natural logarithmic function.2. \( \sqrt{t} \) - a square root function.Our task is to find the derivative of this function with respect to \( t \).

Step 2: Apply the Logarithm Power Rule

Rewrite \( \ln(t^{3/2}) \) by using the logarithm power rule: \( \ln(a^b) = b \cdot \ln(a) \). Thus, we have:\[ \ln(t^{3/2}) = \frac{3}{2} \cdot \ln(t) \].

Step 3: Differentiate the Logarithmic Term

Differentiate \( \frac{3}{2} \cdot \ln(t) \) with respect to \( t \):The derivative of \( \ln(t) \) with respect to \( t \) is \( \frac{1}{t} \). Therefore:\[ \frac{d}{dt} \left( \frac{3}{2} \cdot \ln(t) \right) = \frac{3}{2} \cdot \frac{1}{t} = \frac{3}{2t} \].

Step 4: Differentiate the Square Root Term

The square root term \( \sqrt{t} \) can be rewritten as \( t^{1/2} \). Using the power rule for differentiation, \( \frac{d}{dt}(t^n) = n \cdot t^{n-1} \), we find:\[ \frac{d}{dt}(t^{1/2}) = \frac{1}{2} t^{-1/2} = \frac{1}{2\sqrt{t}} \].

Step 5: Combine Derivatives

Add the derivatives of each term to find the derivative of the entire function \( y \):\[ \frac{dy}{dt} = \frac{3}{2t} + \frac{1}{2\sqrt{t}} \].

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful tool for finding derivatives of functions that are products or quotients of multiple terms, especially when these terms involve exponentials or powers. In this exercise, we apply logarithmic differentiation to the term \( \ln(t^{3/2}) \). By utilizing the logarithm power rule, we can rewrite the expression to make differentiation straightforward. The rule states that \( \ln(a^b) = b \cdot \ln(a) \). This allows us to transform \( \ln(t^{3/2}) \) into \( \frac{3}{2} \cdot \ln(t) \). This transformation simplifies the differentiation process as now we only need to differentiate a linear multiple of a natural log function instead of a more complex expression.

Power Rule

The power rule for differentiation is an essential tool in calculus. It helps us compute the derivative of any function of the form \( t^n \). The rule is rather simple: if \( y = t^n \), then the derivative, \( \frac{dy}{dt} \), is \( n \cdot t^{n-1} \). In this problem, we use the power rule to differentiate \( \sqrt{t} \). By rewriting \( \sqrt{t} \) as \( t^{1/2} \), we can apply the power rule easily. This gives us a derivative of \( \frac{1}{2} t^{-1/2} \), which can also be expressed as \( \frac{1}{2\sqrt{t}} \). The power rule makes it easy to handle exponents in functions, especially when they are simple roots or powers.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with base \( e \), where \( e \approx 2.71828 \), a remarkable mathematical constant. It is commonly used in calculus due to its well-understood properties and its relationship with exponential functions.

• One key derivative rule involving natural logarithms is that \( \frac{d}{dt}[\ln(t)] = \frac{1}{t} \).
• This property becomes particularly useful when working with complex logarithmic functions.

In the exercise, we use the derivative property of natural log to elucidate the derivative of \( \frac{3}{2} \cdot \ln(t) \). This component is simplified by recognizing that the derivative of the whole term is just \( \frac{3}{2} \cdot \frac{1}{t} = \frac{3}{2t} \). The natural logarithm's derivative property is central to simplifying the process, allowing us to tackle expressions that involve natural log.

Square Root Function

The square root function is another fundamental concept in calculus. It's represented as \( \sqrt{t} \), which can also be written as \( t^{1/2} \). Rewriting square roots as exponentials allows the application of the power rule. Differentiating square roots directly is not feasible without exponent conversion.

• To solve our exercise, we first convert \( \sqrt{t} \) into \( t^{1/2} \),
• and then apply the power rule to find its derivative.

We determine that the derivative is \( \frac{1}{2\sqrt{t}} \) or \( \frac{1}{2} t^{-1/2} \). Understanding how to change forms of expressions such as square roots into powers can simplify derivatives of more complex structures. This knowledge allows for straightforward application of the power rule, giving us insight into the behavior of functions involving square roots.