Problem 6 - Single Variable Calculus

Question

Find the derivative of the function. $$ g(v)=e^{v} \sin v-2 v \csc v $$

Step-by-Step Solution

Answer

The short answer is: \[ g'(v) = e^v \sin v + e^v \cos v - 2 - \csc v \cot v \]

Step 1: Differentiate term 1 (Product Rule)

To differentiate the first term, we apply the product rule since it is the product of two functions: \(\frac{d}{dv}(e^v \sin v) = e^v \sin v + e^v \cos v\), where the first term is the product of the derivative of \(e^v \) and the original sine function, and the second term is the product of the original \(e^v\) term and the derivative of the sine function.

Step 2: Differentiate term 2 (Derivative of a Constant)

The second term in the given function is a constant multiplied by a variable. The derivative of a constant times a variable is just that constant, so: \(\frac{d}{dv}(-2v) = -2\)

Step 3: Differentiate term 3 (Derivative of Cosecant)

The third term is the cosecant function, which is the reciprocal of the sine function. The derivative of the cosecant is given by: \(\frac{d}{dv}(\csc v) = -\csc v \cot v\)

Step 4: Combine the Results

Now, we need to combine the results obtained in Steps 1, 2, and 3 to find the overall derivative of the function. \[ g'(v) = \frac{d}{dv}(e^v \sin v - 2v \csc v) = e^v \sin v + e^v \cos v - 2 - \csc v \cot v \] Thus, the derivative of the given function is: \[ g'(v) = e^v \sin v + e^v \cos v - 2 - \csc v \cot v \]

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of the product of two functions. If you have two functions, say \( f(x) \) and \( g(x) \), and you want to find the derivative of their product \( f(x) \cdot g(x) \), the product rule comes into play. It states that the derivative of this product is expressed as: \[ \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \]This means you take the derivative of the first function \( f(x) \), multiply it by the second function \( g(x) \) as it is, then add the first function \( f(x) \) untouched multiplied by the derivative of the second function \( g(x) \).

• For example, to differentiate \( e^v \sin v \) using the product rule, we take the derivative of \( e^v \) which is still \( e^v \) since the exponential function is self-derivative, and the derivative of \( \sin v \) which is \( \cos v \).
• So the differentiation of \( e^v \sin v \) is: \( e^v \sin v + e^v \cos v \).
• This results in the added expression combining these derivatives.

Understanding this rule is key because it simplifies the process of differentiating many complex expressions involving products.

Derivative of Exponential Functions

Exponential functions, commonly denoted as \( e^x \), have unique properties when it comes to differentiation. The derivative of the exponential function with respect to its variable is noteworthy because it results in the same function back.

• The general formula is: \( \frac{d}{dx}[e^x] = e^x \).
• This means that as you differentiate \( e^x \) with respect to \( x \), nothing about the function changes – it remains \( e^x \).

This self-derivative nature makes exponential functions straightforward to work with in calculus.

• When dealing with functions that combine exponential components, such as \( e^v \sin v \), knowing this derivative facilitates the application of other rules, such as the product rule, without overly complicating the expression.

Whether you're tackling one term or multiple expressions involving \( e^x \), remembering its derivative property will save you time and effort.

Trigonometric Derivatives

Trigonometric derivatives are essential when dealing with functions involving sine, cosine, tangent, and other trigonometric attributes. Each trigonometric function has its specific derivative.

• The derivative of \( \sin v \) is \( \cos v \).
• The derivative of \( \cos v \) is \( -\sin v \).
• The derivative of \( \csc v \) (cosecant) is particularly interesting: \( -\csc v \cot v \).

These rules are crucial when working through differentiation processes involving trigonometric terms.

• In the context of the function \( g(v) = e^v \sin v - 2v \csc v \), understanding the derivatives of \( \sin v \) and \( \csc v \) allows you to apply and combine rules like the product rule effectively.
• Using these trigonometric derivatives, you can dissect each term's behavior concerning the variable and find the overall derivative of combined functions.

This strengthens your ability to handle varied functions and proofs in calculus with confidence and accuracy.