Problem 49 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$y=\cos \left(e^{-\theta^{2}}\right)$$

Step-by-Step Solution

Answer

The derivative is \( 2\theta e^{-\theta^2} \sin(e^{-\theta^2}) \).

Step 1: Identify the Function Type

The given function is a composition of functions: a cosine function and an exponential function inside the cosine. This requires the use of the chain rule.

Step 2: Apply the Chain Rule

To find the derivative of the composite function, we apply the chain rule. The chain rule states that if you have a function inside another function, \( f(g(x)) \), then the derivative is \( f'(g(x)) \cdot g'(x) \). Here, the outer function is \( \cos(u) \) where \( u = e^{- heta^2} \).

Step 3: Derivative of the Outer Function

First, take the derivative of the outer function, \( \cos(u) \), which is \( -\sin(u) \). So, \( \frac{d}{du}[\cos(u)] = -\sin(u) \).

Step 4: Derivative of the Inner Function

Next, find the derivative of the inner function, \( e^{-\theta^2} \). Use the chain rule again here: the derivative of \( e^{x} \) is \( e^{x} \), multiplied by the derivative of \( -\theta^2 \), which is \( -2\theta \). So, \( \frac{d}{d\theta}[e^{-\theta^2}] = e^{-\theta^2} \cdot (-2\theta) = -2\theta e^{-\theta^2} \).

Step 5: Combine the Derivatives

Multiply the derivative of the outer function by the derivative of the inner function as per the chain rule. The derivative \( \frac{dy}{d\theta} \) becomes \(-\sin(e^{-\theta^2}) \cdot (-2\theta e^{-\theta^2}) \).

Step 6: Simplify the Expression

Simplify the expression by multiplying the terms: \(-\sin(e^{-\theta^2}) \cdot (-2\theta e^{-\theta^2}) = 2\theta e^{-\theta^2} \sin(e^{-\theta^2}) \). This is as simplified as the expression will get.

Key Concepts

Understanding Derivatives

Derivatives are fundamental in calculus and represent the rate of change of a function with respect to a variable. When you see a function like \(y = \cos \left(e^{-\theta^{2}}\right)\), finding its derivative means determining how \(y\) changes as the variable \(\theta\) changes. This involves some rules and techniques of calculus because the components of this function are not straightforward, just like a simple polynomial.

There are several rules in calculus that guide us in finding derivatives:

• Power Rule: Used for functions with powers of \(x\), such as \(x^n\).
• Product Rule: Used when two functions are multiplied together.
• Quotient Rule: Used for dividing functions.
• Chain Rule: Key for composite functions, as discussed here.

By mastering these rules, students can tackle a wide range of derivative problems effectively.

The Nature of Composite Functions

Composite functions occur when one function is nested within another, such as \(f(g(x))\). In our given example, we have the function \(y = \cos(e^{-\theta^2})\). This means the cosine function encapsulates the function \(e^{-\theta^2}\), making it a composite function.

Understanding composites requires us to breakdown each function within the composite:

• The outer function here is \(\cos(u)\), encapsulating another.
• The inner function is \(u = e^{-\theta^2}\), nestled inside.

The chain rule is essential for differentiating composites. It allows us to find the derivative of these layered functions by dealing with each layer step by step, starting from the outside function and working inwards.

The Role of Calculus in Problem Solving

Calculus is the branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It’s a vast field essential for understanding change and motion. When dealing with problems like \(y = \cos(e^{-\theta^{2}})\), calculus helps provide solutions by leveraging the concepts of derivatives and chain rule.

Calculus can often feel abstract, but it offers great power:

• It allows us to predict behavior (like motion) over time.
• It’s crucial in fields from physics (tracking the trajectory of a ball) to economics (finding cost functions).
• It utilizes tools like derivative rules and integrals to solve complex problems.

For students, grasping these concepts opens the door to understanding complex systems and provides a framework to solve real-world problems. Using calculus, they learn not only to handle mathematical expressions but also to know why and how these expressions apply to daily life and natural phenomena.