Problem 22 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=e^{\sin t}\left(\ln t^{2}+1\right)$$

Step-by-Step Solution

Answer

The derivative of the function is \(y' = e^{\sin t} \cdot \cos t \cdot \ln(t^2) + e^{\sin t} \cdot \cos t + \frac{2e^{\sin t}}{t}\.\)

Step 1: Identify the Rule for Differentiation

The function is composed of a product of two functions: \( e^{\sin t} \) and \( \ln(t^2) + 1 \). We'll use the product rule to differentiate, which is \( (uv)' = u'v + uv' \), where \( u = e^{\sin t} \) and \( v = \ln(t^2) + 1 \).

Step 2: Differentiate \( u = e^{\sin t} \)

To differentiate \( u = e^{\sin t} \), we use the chain rule. This gives us \( u' = e^{\sin t} \cdot \cos t \), since the derivative of \( e^x \) is \( e^x \) and the derivative of \( \sin t \) is \( \cos t \).

Step 3: Differentiate \( v = \ln(t^2) + 1 \)

\( v = \ln(t^2) + 1 \) can be simplified to \( v = 2\ln(t) + 1 \) using the logarithmic identity \( \ln(a^b) = b\ln(a) \). The derivative of \( 2\ln(t) \) is \( \frac{2}{t} \), and the derivative of a constant (1) is 0. Thus, \( v' = \frac{2}{t} \).

Step 4: Apply the Product Rule

Using the product rule \( (uv)' = u'v + uv' \), we substitute the derivatives found previously:- \( u' = e^{\sin t} \cdot \cos t \)- \( v' = \frac{2}{t} \)- \( u = e^{\sin t} \)- \( v = \ln(t^2) + 1 \)The derivative \( y' = (e^{\sin t} \cdot \cos t)(\ln(t^2) + 1) + (e^{\sin t})(\frac{2}{t}) \).

Step 5: Simplify the Expression

The derivative is\[y' = e^{\sin t} \cdot \cos t \cdot \ln(t^2) + e^{\sin t} \cdot \cos t + \frac{2e^{\sin t}}{t}.\]This expression cannot be simplified further significantly, so it is our final result.

Key Concepts

Understanding the Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of composite functions. When a function is composed of two or more functions, the chain rule helps us find the rate at which the output of the entire system changes.
Here's the basic idea: if you have a composition of functions such as \( f(g(x)) \), the derivative is given by \( f'(g(x)) \times g'(x) \). In simple terms, you find the derivative of the outer function while keeping the inner function the same, and multiply it by the derivative of the inner function.
In our exercise, we applied the chain rule to \( e^{\sin t} \). The outer function is \( e^x \) and the inner function is \( \sin t \). By using the chain rule, we found that the derivative is \( e^{\sin t} \cdot \cos t \). This crucial step allows us to differentiate composite parts of functions effectively, especially when they involve trigonometric or exponential expressions.

Exploring the Product Rule

The product rule is a core concept in differentiating functions that are multiplied together. It's used when you need to find the derivative of a product of two functions, say \( u(x) \) and \( v(x) \). The formula is: \( (uv)' = u'v + uv' \).
This rule ensures that both functions in the product are accounted for when differentiating the product as a whole.
In our example, we were tasked with differentiating \( e^{\sin t} \cdot (\ln(t^2) + 1) \). By identifying \( u(t) = e^{\sin t} \) and \( v(t) = \ln(t^2) + 1 \), we applied the product rule. We first found the derivatives of \( u \) and \( v \), then used the product rule formula to find the overall derivative. Each part contributes to the final result, accounting for changes in both functions at once.

Mastering Logarithmic Differentiation

Logarithmic differentiation is a technique often employed when differentiating functions that involve logarithms, especially when dealing with powers and products. Utilizing properties of logarithms can simplify the differentiation process.
In our problem, we faced a logarithmic expression \( \ln(t^2) \). Using the identity for logarithmic powers \( \ln(a^b) = b \ln(a) \), we simplified \( \ln(t^2) \) to \( 2\ln(t) \), which made it much easier to differentiate.

• This identity allowed us to transform a more complex logarithmic expression into a simpler, linear form.
• Once simplified, the differentiation becomes straightforward, as \( \ln(t) \) has a derivative of \( \frac{1}{t} \). Thus, \( 2\ln(t) \) has a derivative of \( \frac{2}{t} \).

This strategy is particularly useful when dealing with products and quotients or any function where direct differentiation seems complex. Logarithmic differentiation gives a new perspective and simplification avenue when tackling intricate functions.

Introduction to Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It is essential for modeling and understanding changes and dynamics within various systems.
At its core, calculus provides tools for analyzing continuous change and is broadly divided into two main parts: differential calculus and integral calculus. Differential calculus deals with the concept of a derivative, which represents an instantaneous rate of change. Meanwhile, integral calculus focuses on accumulation of quantities and the area under curves.
In our exercise, the focus was on finding the derivative of a given function. The derivative gives a precise measure of how one quantity changes in relation to another. Through understanding and applying calculus concepts like the chain rule, product rule, and logarithmic differentiation, we can tackle complex differentiation problems efficiently. Calculus is not just theoretical; it's highly practical and applicable in physics, engineering, economics, statistics, and beyond, anytime you need to measure change.