Problem 13 - Calculus - AP Edition

Question

Find \(f^{\prime}(x)\) $$f(x)=\sqrt{4+\sqrt{3 x}}$$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{3}{4\sqrt{3x}\sqrt{4 + \sqrt{3x}}} \).

Step 1: Recognize the Chain Rule

The function given is a composite function, meaning it can be broken down into two or more simpler functions. The outer function is the square root, and the inner function is another square root. The chain rule in calculus states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

Step 2: Differentiate the Outer Function

Consider the outer function to be \( g(u) = \sqrt{u} \). The derivative of \( g(u) \) with respect to \( u \) is \( g'(u) = \frac{1}{2\sqrt{u}} \). In our case, \( u = 4 + \sqrt{3x} \), so the derivative of the outer function evaluated at the inner function is \( \frac{1}{2\sqrt{4 + \sqrt{3x}}} \).

Step 3: Differentiate the Inner Function

The inner function is \( h(x) = 4 + \sqrt{3x} \). We need to find its derivative. The first part, 4, is a constant, so its derivative is 0. For the second part, \( \sqrt{3x} \), it's a composition of a square root and a linear function, so its derivative is \( \frac{d}{dx} (3x)^{1/2} = \frac{1}{2}(3x)^{-1/2} \times 3 \). This simplifies to \( \frac{3}{2\sqrt{3x}} \).

Step 4: Apply the Chain Rule

Now we combine the results from Steps 2 and 3 using the chain rule. Substitute the derivatives into the chain rule formula: \( f'(x) = g'(h(x)) \cdot h'(x) \). This results in: \[ f'(x) = \frac{1}{2\sqrt{4 + \sqrt{3x}}} \times \frac{3}{2\sqrt{3x}} = \frac{3}{4\sqrt{3x}\sqrt{4 + \sqrt{3x}}} \].

Key Concepts

Composite Function

In calculus, a composite function combines two or more functions into a single function. Here, our given function is composite because it involves a "function within a function." Specifically, it's a series of square roots nested inside each other. For the function \( f(x) = \sqrt{4 + \sqrt{3x}} \), we can identify two key layers:

• The outer function: \( g(u) = \sqrt{u} \)
• The inner function: \( h(x) = 4 + \sqrt{3x} \)

This can be thought of as plugging one function into another, similar to a manufacturing process where you input one part to get a modified output. Understanding the breakdown of composite functions is crucial because it allows us to apply the chain rule effectively in differentiation.

Derivative

Derivatives measure how a function changes as its input changes. It's like asking how fast you're moving at any given moment if you were looking at a position-time graph. For composite functions, we differentiate layer by layer.The function \( f(x) = \sqrt{4 + \sqrt{3x}} \) needs its outer and inner functions differentiated separately:

• Outer Derivative: If your outer function is \( g(u) = \sqrt{u} \), its derivative is \( g'(u) = \frac{1}{2\sqrt{u}} \).
• Inner Derivative: For the inner function \( h(x) = 4 + \sqrt{3x} \), the derivative of \( \sqrt{3x} \) is \( \frac{3}{2\sqrt{3x}} \), since the derivative of \( x^n \) is \( nx^{n-1} \).

These derivatives unravel the layers to see how changes affect the whole function. Each derivative reveals how sensitive the function is to changes in its variables.

Calculus Steps

Differentiating composite functions requires the chain rule, a crucial method in calculus. The chain rule lets us differentiate step-by-step to capture every layer in a composite function.For the problem \( f(x) = \sqrt{4 + \sqrt{3x}} \), we applied these calculus steps:

• First, identify the layers of functions involved (outer and inner).
• Differentiate the outer function using its derivative formula.
• Proceed to differentiate any inner function(s) present.
• Finally, the chain rule combines these derivatives: \( f'(x) = g'(h(x)) \cdot h'(x) \).

In this exercise, the result is a single derivative expression: \[f'(x) = \frac{3}{4\sqrt{3x}\sqrt{4 + \sqrt{3x}}}\]Using these calculus steps systematically ensures accuracy and understanding in differentiation processes, especially for complex functions.