Problem 24 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ y=e^{-2 t} \cos 4 t $$

Step-by-Step Solution

Answer

The derivative is \( e^{-2t}(-2\cos 4t - 4\sin 4t) \).

Step 1: Identify the Functions to Differentiate

The function \( y = e^{-2t} \cos 4t \) is a product of two functions: \( e^{-2t} \) and \( \cos 4t \). This means we will apply the product rule for differentiation.

Step 2: Recall the Product Rule

The product rule states that if you have two functions, \( u \) and \( v \), then the derivative is \( \frac{d}{dt}(uv) = u'v + uv' \). In this case, \( u = e^{-2t} \) and \( v = \cos 4t \).

Step 3: Differentiate each Function Separately

First, find the derivative of \( u = e^{-2t} \):\( u' = \frac{d}{dt}(e^{-2t}) = -2e^{-2t} \) using the chain rule.Next, find the derivative of \( v = \cos 4t \):\( v' = \frac{d}{dt}(\cos 4t) = -4\sin 4t \) again using the chain rule.

Step 4: Apply the Product Rule

Using the derivative rules from Step 2, substitute the derivatives found in Step 3:\( \frac{d}{dt}(y) = u'v + uv' = (-2e^{-2t})(\cos 4t) + (e^{-2t})(-4\sin 4t) \).

Step 5: Simplify the Expression

Simplify the expression obtained in Step 4: \( \frac{d}{dt}(y) = -2e^{-2t} \cos 4t - 4e^{-2t} \sin 4t \).Factor \( e^{-2t} \) out of the expression:\( \frac{d}{dt}(y) = e^{-2t}(-2\cos 4t - 4\sin 4t) \).

Key Concepts

Product Rule

When dealing with the differentiation of a function that is a product of two other functions, like in the exercise, we apply the product rule. The product rule's purpose is to provide a method for finding the derivative of a product of two functions. If you have functions \( u(t) \) and \( v(t) \), their product \( y(t) = u(t)v(t) \) has a derivative given by \( y'(t) = u'(t)v(t) + u(t)v'(t) \).
This rule helps us break down complexity by allowing us to treat each function separately. In our example, \( u(t) = e^{-2t} \) and \( v(t) = \cos 4t \).
We break the problem into smaller parts, first finding the derivatives \( u'(t) \) and \( v'(t) \), and then applying the rule.

• The derivative of the first function, \( e^{-2t} \), is found using the chain rule, resulting in \( u'(t) = -2e^{-2t} \).
• The derivative of the second function, \( \cos 4t \), is \( v'(t) = -4\sin 4t \), also found using the chain rule.
• Substituting these derivatives back into the product rule gives us the overall derivative.

Understanding the product rule is crucial because it simplifies the process of differentiation for products of functions.

Chain Rule

The chain rule is one of the fundamental rules for differentiation and is instrumental in solving the given problem. It's used when differentiating composite functions, where one function is nested inside another.
If you have a function \( y = f(g(x)) \), then the derivative is given by \( y' = f'(g(x))g'(x) \).
This allows you to differentiate each layer of the function separately.
In our exercise, the chain rule is utilized twice:

• Firstly, for \( e^{-2t} \), recognize that \( f(x) = e^x \) and \( g(t) = -2t \), hence the derivative \( u' = (-2)e^{-2t} \).
• Secondly, for \( \cos 4t \), decompose it into \( f(x) = \cos x \) and \( g(t) = 4t \), leading to the derivative \( v' = (-4)\sin 4t \).

Each step of the chain rule involves taking the derivative of the outer function and multiplying it by the derivative of the inner function, providing a powerful tool for calculating complex derivatives.

Exponential Functions

Exponential functions are a key mathematical concept with unique properties. They take the form \( y = e^{u(t)} \), where \( e \) is a constant approximately equal to 2.718.
Such functions are notable for their rapid growth or decay and are frequently encountered in differentiation problems due to their inherent properties.
The derivative of an exponential function \( e^{u(t)} \) is straightforward to compute using the chain rule. It results in \( u'(t) e^{u(t)} \).
This is because the derivative of \( e^x \) with respect to \( x \) is \( e^x \), and when applying the chain rule, it requires an additional multiplication by the derivative of the exponent.

• In the exercise, the function \( e^{-2t} \) is differentiated to \( -2e^{-2t} \), demonstrating this principle.
• This derivative is employed in the product rule as part of the overall derivative computation.
• The property that exponential functions remain largely unchanged, apart from a constant factor \( u'(t) \), greatly simplifies calculations.

Consequently, understanding how to handle exponential functions systematically eases the differentiation process of complex expressions.