Problem 61 - University Calculus: Early Transcendentals

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

\(\frac{dy}{dt} = e^{(\cos t + \ln t)} \cdot \left(-\sin t + \frac{1}{t}\right)\).

Step 1: Identify the function and variables

The given function is \(y = e^{(\cos t + \ln t)}\). Thus, the variable with respect to which we will differentiate is \(t\).

Step 2: Recall the chain rule for differentiation

The chain rule states that if you have a composite function \(y = e^{u(t)}\), where \(u(t)\) is another function of \(t\), then the derivative is \(\frac{dy}{dt} = e^{u(t)} \cdot \frac{du}{dt}\).

Step 3: Differentiate the exponent function

Identify \(u(t) = \cos t + \ln t\). Thus, the next task is to find \(\frac{du}{dt}\).

Step 4: Find the derivative of \(u(t) = \cos t + \ln t\)

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

Step 5: Apply the chain rule

Now that we have \(u(t)\) and \(\frac{du}{dt} = -\sin t + \frac{1}{t}\), we use the chain rule: \[ \frac{dy}{dt} = e^{(\cos t + \ln t)} \cdot (-\sin t + \frac{1}{t}). \]

Step 6: State the final derivative

Substitute back the calculated derivative into the chain rule result: \[ \frac{dy}{dt} = e^{(\cos t + \ln t)} \cdot \left(-\sin t + \frac{1}{t}\right). \] This is the derivative of \(y\) with respect to \(t\).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. If you have a function defined as a combination, such as \(y = e^{u(t)}\), then it can be quite handy. Here, \(u(t)\) is a function of \(t\) itself. The chain rule suggests that the derivative of \(y\) with respect to \(t\) is given by the product of the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

• This means \(\frac{dy}{dt} = e^{u(t)} \cdot \frac{du}{dt}\).

Using the chain rule allows you to break down complicated differentiation tasks into more manageable parts. First, differentiate the outer layer, keeping the inside the same, and then multiply it by the derivative of the inside.

Exponential Functions

Exponential functions have the form \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718. These functions have unique properties that make them vital in various fields, especially calculus. The derivative of an exponential function is quite special—it's the same exponential function itself, multiplied by the derivative of its exponent.

• In our exercise, the function is \(y = e^{\cos t + \ln t}\), and according to our rules, the derivative would initially look like the original, \(e^{\cos t + \ln t}\).

This feature makes working with exponential functions smoother because you consistently apply similar rules.

Differentiation Techniques

Differentiation is the process of finding the rate at which a function is changing at any point. There are various techniques for differentiation, which depend on the type of function. In this exercise, some basic differentiation techniques are utilized:

• Derivative of cosine function: The derivative of \(\cos t\) is \(-\sin t\).
• Derivative of natural logarithm: The derivative of \(\ln t\) is \(\frac{1}{t}\).

When dealing with more complex functions, like the sum \(\cos t + \ln t\), each part is differentiated separately and then combined. Making use of these basic derivatives and rules, such as the chain rule, simplifies the process of handling intricate problems.