Problem 66 - Single Variable Calculus

Question

Find the derivative of the function. $$ h(\theta)=\tan ^{-1}\left(\frac{\cos \theta}{2}\right) $$

Step-by-Step Solution

Answer

The derivative of the function \(h(\theta) = \tan^{-1}\left(\frac{\cos\theta}{2}\right)\) is: \[\frac{dh}{d\theta} = -\frac{4\sin\theta}{8 + \cos^2\theta}\]

Step 1: Identify the outer and inner functions

In this problem, our outer function is $$f(u) = \tan^{-1}(u)$$ and our inner function is $$u(\theta) = \frac{\cos\theta}{2}$$.

Step 2: Find the derivative of the outer function

We need to find the derivative of the outer function with respect to its variable, $$u$$. The derivative of the arctangent function is given by: $$ \frac{d}{du}\left(\tan^{-1}(u)\right) = \frac{1}{1 + u^2} $$

Step 3: Find the derivative of the inner function

Now we need to find the derivative of the inner function with respect to its variable, $$\theta$$. The derivative of $$\frac{\cos\theta}{2}$$ is: $$ \frac{d}{d\theta}\left(\frac{\cos\theta}{2}\right) = -\frac{1}{2}\sin\theta $$

Step 4: Apply the chain rule

Finally, we apply the chain rule by multiplying the two derivatives we found in steps 2 and 3. So, the derivative of $$h(\theta)$$ is: $$ \frac{dh}{d\theta} = \left(\frac{1}{1 + \left(\frac{\cos\theta}{2}\right)^2}\right) \cdot (-\frac{1}{2}\sin\theta) $$

Step 5: Simplify the expression

Now, we will simplify the expression, combining the terms: $$ \frac{dh}{d\theta} = -\frac{\sin\theta}{2\left(1 + \frac{\cos^2\theta}{4}\right)} $$ To get rid of the fraction in the denominator, we can multiply both the numerator and the denominator by 4: $$ \frac{dh}{d\theta} = -\frac{4\sin\theta}{8 + \cos^2\theta} $$ The final derivative of the function $$h(\theta) = \tan^{-1}\left(\frac{\cos\theta}{2}\right)$$ is: $$ \frac{dh}{d\theta} = -\frac{4\sin\theta}{8 + \cos^2\theta} $$

Key Concepts

Arctangent Function

The arctangent function, denoted as \( \tan^{-1}(x) \) or \( \arctan(x) \), is an inverse trigonometric function. It gives the angle whose tangent value is \( x \). Understanding this function is key in calculus, especially when dealing with derivatives involving angles and slopes. When you differentiate the arctangent function, you apply a predictable pattern. The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is given by:

• \( \frac{d}{du}\left(\tan^{-1}(u)\right) = \frac{1}{1 + u^2} \)

This expression is essential and used frequently when solving calculus problems involving arctangent functions. In our specific problem, \( u \) is the inner function \( \frac{\cos \theta}{2} \), making it crucial to apply this derivative rule properly. Understanding arctangent and its properties helps in managing complex trigonometric equations in calculus.

Chain Rule

The chain rule is a fundamental calculus tool for computing the derivative of a composition of functions. When you have a function \( h(x) = f(g(x)) \), the derivative \( \frac{dh}{dx} \) is found by multiplying the derivative of the outer function \( f' \) evaluated at the inner function \( g(x) \), by the derivative of the inner function \( g'(x) \). The formula looks like this:

• \( \frac{dh}{dx} = f'(g(x)) \cdot g'(x) \)

In our exercise, the chain rule is crucial because we are dealing with \( h(\theta) = \tan^{-1}\left(\frac{\cos\theta}{2}\right) \). Here, the outer function is the arctangent, and the inner function is \( \frac{\cos \theta}{2} \). Applying the chain rule involves:

• Finding the derivative of the outer function with respect to the inner function.
• Multiplying it by the derivative of the inner function with respect to \( \theta \).

This is how we end up calculating the derivative \( \frac{dh}{d\theta} \) effectively, handling complex nested functions in calculus.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are the building blocks in calculus, especially when dealing with derivatives and integrals. These functions relate to the angles and sides of right triangles, with sine and cosine related to the opposite and adjacent sides over the hypotenuse.

• The derivative of \( \cos \theta \) is \( -\sin \theta \).
• These properties are pivotal when differentiating composed trigonometric functions.

In our derivative problem, knowing these derivatives was essential when differentiating the inner function \( \frac{\cos\theta}{2} \). The derivative was \( -\frac{1}{2}\sin\theta \), which is vital to find after applying the chain rule.Understanding trigonometric identities and their derivatives helps simplify complex calculus problems, especially those involving nested functions like in our original task.

Calculus

Calculus is the branch of mathematics that studies change, primarily focusing on derivatives and integrals. Derivatives help us understand the rate at which one quantity changes with respect to another. This is foundational when finding slopes of tangent lines and rates of change in physical systems.

• In calculus, derivatives of functions like \( \tan^{-1}(x) \), and principles like the chain rule, allow us to tackle problems involving derivatives of complex functions.
• Calculus unlocks the ability to model and solve real-world problems, such as motion, growth, and decay.

In our problem, applying calculus principles, particularly derivative rules, chain rule, and simplification methods, brings us to the final simplified derivative of complex functions. The exercise demonstrates how calculus connects concepts such as inverse trigonometric functions and these principles to derive solutions efficiently.