Problem 5 - Single Variable Calculus

Question

Find the derivative of the function. $$ f(u)=e^{u} \cot u $$

Step-by-Step Solution

Answer

The derivative of \(f(u) = e^u \cot u\) is \(f'(u) = e^u (\cot u - \csc^2 u)\).

Step 1: Identify h(u) and k(u)

We have: - \(h(u) = e^u\) - \(k(u) = \cot u\) Next, we will find the derivatives of both functions.

Step 2: Find the derivatives of h(u) and k(u)

We have: - \(h'(u) = \frac{d}{du}(e^u) = e^u\) - \(k'(u) = \frac{d}{du}(\cot u) = -\csc^2{u}\) Now, we will apply the product rule to find the derivative of \(f(u)\).

Step 3: Apply the product rule

Using the product rule, we have: \(f'(u) = h'(u) \cdot k(u) + h(u) \cdot k'(u)\) Substitute the derivatives of \(h(u)\) and \(k(u)\) that we found in Step 2: \(f'(u) = e^u \cdot \cot u - e^u \cdot \csc^2 u\) Finally, factor out \(e^u\).

Step 4: Write the final answer

Factor out \(e^u\) from the expression: \(f'(u) = e^u (\cot u - \csc^2 u)\) So, the derivative of \(f(u) = e^u \cot u\) is \(f'(u) = e^u (\cot u - \csc^2 u)\).

Key Concepts

Product Rule

The product rule is an essential technique in calculus for finding the derivative of a product of two functions. If you have two functions, say \( h(u) \) and \( k(u) \), and you want to differentiate their product, the product rule provides a straightforward method.

The product rule is stated as:

• Given \( f(u) = h(u) \, k(u) \), then the derivative \( f'(u) \) is \( h'(u) \, k(u) + h(u) \, k'(u) \).

Simply put, you take the derivative of the first function and multiply it by the second function as it is, and then add the product of the first function as it is and the derivative of the second function.

This rule is invaluable, especially when dealing with complex functions that are products of simpler functions, like in our problem where \( f(u) = e^u \cot u \).

Exponential Functions

Exponential functions are a category of mathematical functions in which the variable is in the exponent. The function \( e^u \) is a quintessential example of an exponential function, where \( e \) is Euler's number, approximately equal to 2.71828.

Key properties of exponential functions include:

• They grow at an increasing rate.
• The derivative of \( e^u \) with respect to \( u \) is simply \( e^u \). This unique property makes calculus with exponential functions straightforward.

Because the derivative of an exponential function is proportional to the original function itself, it often features in models of growth, decay, and many natural processes.

In our example, knowing that the derivative of \( e^u \) is \( e^u \) simplifies our calculation when applying the product rule.

Trigonometric Functions

Trigonometric functions are functions related to angles and sides of triangles. They also play vital roles in various fields such as physics and engineering. In our problem, we work with the cotangent function, \( \cot u \).

Important trigonometric derivative identities include:

• The derivative of \( \cot u \) is \( -\csc^2 u \).
• Cosecant, represented as \( \csc u \), is the reciprocal of the sine function.

Understanding these derivatives helps when using the product rule because it simplifies the differentiation process. Knowing that \( \cot u \) becomes \( -\csc^2 u \) is critical for correctly solving the derivative of the given function.

Differentiation Techniques

Differentiation techniques like the product rule, chain rule, and more are used to find the rate of change of functions. Here we focus primarily on the product rule as it was used to find the derivative of \( f(u) = e^u \cot u \).

Steps when differentiating using these techniques generally include:

• Identify the functions and their derivatives involved.
• Apply appropriate rules such as the product rule when functions are multiplied.
• Simplify the expression by combining like terms or factoring where possible, just like we factored \( e^u \) in the final expression.

Mastering these techniques enables you to tackle more complex calculus problems and deepens your understanding of function behavior. Differentiation forms the backbone of calculus, emphasizing the importance of these rules.