Chapter 7: Problem 21
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=e^{(\cos t+\ln t)}$$
Short Answer
Expert verified
The derivative is \( y' = e^{(\cos t + \ln t)} \left(-\sin t + \frac{1}{t}\right) \).
Step by step solution
01
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.
02
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 \( e^u \cdot \frac{du}{dt} \).
03
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, \( \frac{du}{dt} = -\sin t + \frac{1}{t} \).
04
Apply the Chain Rule
Substitute \( \frac{du}{dt} = -\sin t + \frac{1}{t} \) back into the chain rule derivative \( y' = e^{(\cos t + \ln t)} \cdot \left(-\sin t + \frac{1}{t}\right) \).
05
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 \( t \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
- 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 \( t \).
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.
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:
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.
- The derivative of \( \cos t \) is \( -\sin t \).
- The derivative of \( \sin t \) is \( \cos t \).
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.