Problem 21 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

The derivative is \( y' = e^{(\cos t + \ln t)} \left(-\sin t + \frac{1}{t}\right) \).

Step 1: Identify the Function Components

The given function is \( y = e^{(\cos t + \ln t)} \). The exponent consists of two separate functions: \( \cos t \) and \( \ln t \). We will apply the chain rule to find the derivative.

Step 2: Differentiate Using the Chain Rule

To differentiate \( y = e^{u} \), where \( u = \cos t + \ln t \), we use the chain rule. The derivative of \( e^u \) with respect to \( t \) is \( e^u \cdot \frac{du}{dt} \).

Step 3: Differentiate the Inner Function

Find the derivative of the inner function \( u = \cos t + \ln t \). The derivative of \( \cos t \) is \( -\sin t \), and the derivative of \( \ln t \) is \( \frac{1}{t} \). Thus, \( \frac{du}{dt} = -\sin t + \frac{1}{t} \).

Step 4: Apply the Chain Rule

Substitute \( \frac{du}{dt} = -\sin t + \frac{1}{t} \) back into the chain rule derivative \( y' = e^{(\cos t + \ln t)} \cdot \left(-\sin t + \frac{1}{t}\right) \).

Step 5: Simplify the Expression

The expression can be simplified to \( y' = e^{(\cos t + \ln t)} \cdot \left(-\sin t + \frac{1}{t}\right) \). Thus, this is the derivative of the function with respect to \( t \).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus that helps us find the derivative of a composite function. A composite function is essentially a function inside another function, like our example with the exponential function raised to the power of a sum.

• The outer function is the exponential function, represented as \( e^u \).
• The inner function is \( u = \cos t + \ln t \). Here, \( u \) is a function of \( t \).

Using the chain rule, we differentiate the outer function \( e^u \) with respect to \( u \), which gives us \( e^u \).
Next, we multiply by the derivative of the inner function \( u \) with respect to \( t \), denoted as \( \frac{du}{dt} \). This gives us the complete derivative: \( \frac{dy}{dt} = e^u \cdot \frac{du}{dt} \).
Understanding each step ensures you can tackle more complicated derivative problems.

Exploring Exponential Functions

Exponential functions involve the mathematical constant \( e \), approximately equal to 2.71828. The function \( e^x \) is unique because its derivative is itself, meaning \( \frac{d}{dx}(e^x) = e^x \). This property makes it highly valuable in calculus.Moving to our exercise, where the exponential function is \( e^{(\cos t + \ln t)} \):

• The base \( e \) plays a crucial role in maintaining the properties of the function, regardless of the complexity of the exponent \((\cos t + \ln t)\).
• When differentiating, we keep the structure \( e^u \) intact while focusing on the exponent's derivative. This maintains the elegance of the exponential property.

In developmental terms, once we're comfortable with the derivative of \( e^x \), handling more complex scenarios becomes more approachable.

Working with Trigonometric Functions

Trigonometric functions, like \( \cos t \) and \( \sin t \), are common in calculus exercises because they model cyclical behavior. Knowing their derivatives is essential:

• The derivative of \( \cos t \) is \( -\sin t \).
• The derivative of \( \sin t \) is \( \cos t \).

In the given exercise, the term \( \cos t \) appears as a part of the composite function. To differentiate it, we find \( \frac{d}{dt}(\cos t) = -\sin t \).
Communicating effectively between trigonometric and exponential functions is key to simplifying derivative calculations.
Pairing these concepts with the chain rule allows us to navigate even more intricate calculus problems easily.