Problem 55 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=\tan \frac{1}{x} $$

Step-by-Step Solution

Answer

The derivative is \(-\sec^2\left(\frac{1}{x}\right) \cdot \frac{1}{x^2}\).

Step 1: Identify the Composite Function

The given function is a composition of functions: an outer tangent function and an inner reciprocal function. Let's identify these: the outer function is \(u(x) = \tan(x)\) and the inner function is \(v(x) = \frac{1}{x}\). We will need to apply the chain rule to differentiate.

Step 2: Differentiate the Outer Function

First, differentiate the outer function \(u(v) = \tan(v)\) with respect to \(v\). The derivative of \(\tan(v)\) is \(\sec^2(v)\). Thus, we have \( \frac{du}{dv} = \sec^2\left(\frac{1}{x}\right)\).

Step 3: Differentiate the Inner Function

Next, differentiate the inner function \(v(x) = \frac{1}{x}\) with respect to \(x\). The derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\). So, \( \frac{dv}{dx} = -\frac{1}{x^2}\).

Step 4: Apply the Chain Rule

Using the chain rule for differentiation, the derivative of the composite function \( f(x) = \tan\left(\frac{1}{x}\right) \) is the product of the derivatives of the outer and inner functions. Using the chain rule: \[ \frac{df}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = \sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right). \]

Step 5: Simplify the Result

Now simplify the expression: \[ \frac{df}{dx} = -\sec^2\left(\frac{1}{x}\right) \cdot \frac{1}{x^2}. \] This is the simplified form of the derivative of the given function.

Key Concepts

Understanding the Chain Rule

The chain rule is a powerful technique in calculus for differentiating composite functions. When you have a function within a function, you use the chain rule to find the derivative. It allows us to deconstruct complex problems into more manageable steps. For instance, when you differentiate a composite function like \( f(g(x)) \), the chain rule tells us to first differentiate \( f \) with respect to \( g \), and then multiply by the derivative of \( g \) with respect to \( x \).
This is mathematically represented as:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

The wisdom of the chain rule lies in its simplification approach, breaking down functions into smaller, easier tasks. In essence, it accounts for the rate of change in each layer of a function. Understanding this rule well is essential for tackling problems involving multiple nested functions.

Composite Functions Explained

Composite functions are operations where one function is applied to the result of another function. Imagine them as mathematical nesting dolls, where the output of one function becomes the input for the next. For example, if you have a function \( f(x) = \tan\left(\frac{1}{x}\right) \), you have one function, \( \tan(v) \), encasing another function, \( \frac{1}{x} \).
This is defined as:

• Inner Function: \( v(x) = \frac{1}{x} \)
• Outer Function: \( u(v) = \tan(v) \)

Recognizing and dissecting these components properly are pivotal to applying the chain rule effectively. Composite functions frequently appear in calculus, needing individual manipulation of each part for accurate derivative computation. Whether in physics, engineering, or pure mathematics, composite functions enable the handling of complex relationships between variables.

Exploring Trigonometric Derivatives

Trigonometric derivatives are the derivatives of trigonometric functions. They are crucial in calculus, revealing how trigonometric functions change. Each trigonometric function has a specific derivative. Here are a few common derivatives to remember:

• The derivative of \( \sin(x) \) is \( \cos(x) \)
• The derivative of \( \cos(x) \) is \( -\sin(x) \)
• The derivative of \( \tan(x) \) is \( \sec^2(x) \)
• The derivative of \( \sec(x) \) is \( \sec(x)\tan(x) \)

Trigonometric derivatives often play roles in physics and engineering. Knowing these derivatives helps in solving problems where trigonometric functions are involved, such as oscillations or waves. In our example problem, the derivative of the tangent function, \( \tan(v) \), with respect to \( v \), gave us \( \sec^2(v) \), an essential step in applying the chain rule to this composite function.