In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = e^(cost+lnt)

Verified step by step guidance

1. Identify the variable with respect to which you need to differentiate. Since the function is given as \(y = e^{\cos t + \ln t}\), and the expression involves \(t\), we will differentiate with respect to \(t\).
2. Recall the chain rule for differentiation: if \(y = e^{u(t)}\), then \(\frac{dy}{dt} = e^{u(t)} \cdot \frac{du}{dt}\), where \(u(t) = \cos t + \ln t\) in this case.
3. Find the derivative of the exponent function \(u(t) = \cos t + \ln t\). Differentiate each term separately: \(\frac{d}{dt}(\cos t) = -\sin t\) and \(\frac{d}{dt}(\ln t) = \frac{1}{t}\).
4. Combine the derivatives of the terms to get \(\frac{du}{dt} = -\sin t + \frac{1}{t}\).
5. Write the final expression for the derivative using the chain rule: \(\frac{dy}{dt} = e^{\cos t + \ln t} \cdot \left(-\sin t + \frac{1}{t}\right)\).

Key Concepts

Chain Rule

The chain rule is used to differentiate composite functions. It states that the derivative of a function composed of another function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This is essential when differentiating expressions like e^(cos t + ln t).

Derivatives of Exponential Functions

The derivative of an exponential function e^u, where u is a function of the variable, is e^u multiplied by the derivative of u. Recognizing this helps in differentiating y = e^(cos t + ln t) by first identifying the exponent as a function and then applying the chain rule.

Derivatives of Trigonometric and Logarithmic Functions

Knowing the derivatives of cos t and ln t is crucial. The derivative of cos t with respect to t is -sin t, and the derivative of ln t is 1/t. These derivatives are used when applying the chain rule to find the derivative of the exponent in the given function.