Problem 16 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\tan \sqrt{x} $$

Step-by-Step Solution

Answer

The derivative is \(f'(x) = \frac{\sec^2(\sqrt{x})}{2\sqrt{x}}\).

Step 1: Understand the Problem

We need to find the derivative, \(f'(x)\), of the function \(f(x) = \tan \sqrt{x}\). This requires the use of both the chain rule and the derivative of the tangent function.

Step 2: Apply the Chain Rule

The chain rule tells us that if a function \(g(x)\) can be written as a composition of functions, \(u(v(x))\), then \(g'(x) = u'(v(x)) \cdot v'(x)\). For \(f(x) = \tan(\sqrt{x})\), we identify two functions: the outer function \(u(t) = \tan(t)\) and the inner function \(v(x) = \sqrt{x}\).

Step 3: Derivative of the Outer Function

Find the derivative of the outer function \(u(t) = \tan(t)\). The derivative \(u'(t)\) is \(\sec^2(t)\).

Step 4: Derivative of the Inner Function

Find the derivative of the inner function \(v(x) = \sqrt{x}\). We rewrite this as \(x^{1/2}\) and the derivative \(v'(x)\) is \(\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\).

Step 5: Combine Using the Chain Rule

Now use the chain rule to combine the derivatives. For \(f(x) = \tan(\sqrt{x})\), we apply \(f'(x) = u'(v(x)) \cdot v'(x)\). Substitute \(u'(t) = \sec^2(t)\) evaluated at \(t=\sqrt{x}\) and \(v'(x) = \frac{1}{2\sqrt{x}}\). This gives us \(f'(x) = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\).

Step 6: Simplify the Result

Simplify the expression for the derivative: \(f'(x) = \frac{\sec^2(\sqrt{x})}{2\sqrt{x}}\). Since \(\sec^2(t) = \frac{1}{\cos^2(t)}\), recall this formula if further simplification is needed, but the expression is already simplified.

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. It is the concept of finding the slope of the tangent line to a curve at a given point. Think of it as a way to measure the rate of change. When you have a function like \(f(x) = \tan \sqrt{x}\), you're trying to find its derivative, \(f'(x)\), which tells us how the function's output changes with respect to changes in \(x\).

Derivatives are foundational in calculus because they help us understand the behavior of functions beyond static values. They allow us to predict trends and analyze varying phenomena in more dynamic terms. The process involves different rules and techniques, including the chain rule for composite functions.

Chain Rule

The chain rule is a critical concept when dealing with composite functions, or functions within functions. It provides a way to compute derivatives of such composite functions. The basic idea is simple: differentiate the outer function, and then multiply by the derivative of the inner function.

For example, with the function \(f(x) = \tan(\sqrt{x})\), you need to first identify its components:

• The outer function \(u(t) = \tan(t)\)
• The inner function \(v(x) = \sqrt{x}\)

This means that the derivative \(f'(x)\) is found by calculating \(u'(v(x)) \cdot v'(x)\). First, differentiate the outer function to get \( \sec^2(t)\), and then differentiate the inner function to get \( \frac{1}{2\sqrt{x}}\).

Combining these, the chain rule lets you find that \(f'(x) = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\), illustrating how this powerful tool helps manage complicated calculations efficiently.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are fundamental components in mathematics. They arise from considerations of angles and triangles, and are particularly useful in calculus for modeling periodic phenomena.

In calculus, each trigonometric function has its own derivative:

• \(\frac{d}{dx} \sin(x) = \cos(x)\)
• \(\frac{d}{dx} \cos(x) = -\sin(x)\)
• \(\frac{d}{dx} \tan(x) = \sec^2(x)\)

For \(f(x) = \tan(\sqrt{x})\), understanding the derivative of \(\tan(x)\) as \(\sec^2(x)\) is crucial. This fixes how the entire function \(\tan(\sqrt{x})\) will change, given that \(x\) changes, due to the trigonometric relationship.

Trigonometric functions often introduce complexity in calculus problems, particularly when they are visited within composite functions, demanding a good grasp of differentiation rules, such as the chain rule, to resolve them.