Problem 25 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. $$y=\arctan (\cos \theta)$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{d\theta} = \frac{-\sin \theta}{1+\cos^2 \theta} \).

Step 1: Identify the Function and Variables

The function given is \( y = \arctan(\cos \theta) \). This function is a composition of functions, which requires the use of the chain rule to find the derivative.

Step 2: Apply the Chain Rule

To find the derivative of a composite function like \( \arctan(u) \), where \( u = \cos \theta \), we first need the derivative of the outer function. The derivative of \( \arctan(u) \) is \( \frac{1}{1+u^2} \). Thus, apply this to \( u = \cos \theta \), so it becomes \( \frac{1}{1+(\cos \theta)^2} \).

Step 3: Differentiate the Inner Function

Now differentiate the inner function, \( \cos \theta \), with respect to \( \theta \). The derivative of \( \cos \theta \) is \( -\sin \theta \).

Step 4: Combine Using the Chain Rule

Multiply the derivative of the outer function by the derivative of the inner function. So, the overall derivative \( \frac{dy}{d\theta} \) is: \[ \frac{dy}{d\theta} = \frac{1}{1+(\cos \theta)^2} \times (-\sin \theta) = \frac{-\sin \theta}{1+\cos^2 \theta} \].

Step 5: Simplify the Expression

The expression \( \frac{-\sin \theta}{1+\cos^2 \theta} \) is already quite simplified. However, recognize that \( 1+\cos^2 \theta \) does not simplify directly in this context, so the final derivative is:\[ \frac{dy}{d\theta} = \frac{-\sin \theta}{1+\cos^2 \theta} \].

Key Concepts

Chain Rule

When it comes to derivatives in calculus, the chain rule is a fundamental tool to handle composite functions. This rule helps us find derivatives of functions that are composed of other functions. This rule can be thought of as a way to "peel back" the layers of a composite function and differentiate each piece.

Let's break this down with our example of the function, \( y = \arctan(\cos \theta) \).

• Here, the outer function is \( \arctan(u) \), and
• the inner function is \( u = \cos \theta \).

To apply the chain rule, we first need the derivative of the outer function, \( \arctan(u) \), which is \( \frac{1}{1+u^2} \).
Next, we find the derivative of the inner function, \( \cos \theta \), which is \( -\sin \theta \).
Finally, multiply these derivatives together as per the chain rule formula: the derivative of the outer function with respect to the inner function, multiplied by the derivative of the inner function itself. This results in our final derivative.
Understanding this sequence and relationship between these derivatives is key to mastering problems involving the chain rule.

Trigonometric Functions

Trigonometric functions are a class of functions that are critical in both calculus and geometry. These functions, such as \( \cos \theta \) and \( \sin \theta \), help model rotational and periodic phenomena. Beyond their geometric interpretations, they have a wide range of derivatives that make them interesting and sometimes tricky in calculus.

In our exercise, we are dealing with \( \cos \theta \). The derivative of this trigonometric function is \( -\sin \theta \).

• This negative sign is important because it signifies a phase shift in the periodic wave of our cosine function.
• Each trigonometric function has a specific derivative, and remembering these can aid significantly in calculus problems.

Understanding these derivatives allows for the analysis and interpretation of the rate of change in wave-like phenomena, such as electromagnetic waves or sound waves. The periodic nature of trigonometric functions can also introduce resonance and other characteristics in physical systems.

Composite Functions

Composite functions are functions made by applying one function to the results of another. They are expressed as \( (f \circ g)(x) = f(g(x)) \), where \( f \) and \( g \) are individual functions.

In the context of derivatives, the concept of compositing functions leads directly to the application of the chain rule. This occurs because each layer of function relies on the output of the previous layer.

For the example function \( y = \arctan(\cos \theta) \):

• The function \( \arctan \) is applied to the value from \( \cos \theta \), making our function inherently composite.
• Therefore, the derivative of such a function must respect both the internal (\( \cos \theta \)) and the external (\( \arctan \)) layers.

By recognizing functions as composite, students can apply appropriate rules like the chain rule to successfully find derivatives and solve related calculus problems efficiently.