Question

Evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result. \(f(x)=\frac{1}{\sqrt{x+4}}, \quad\left(0, \frac{1}{2}\right)\)

Short answer

The second derivative of the function \(f(x)=\frac{1}{\sqrt{x+4}}\) at the point \(x=0\) is \(\frac{3}{32}\).

Step-by-step solution

Find the First Derivative of the Function

To start, find the derivative of \(f(x)\), using the chain rule: \[f'(x) = \frac{d}{dx}\left(\frac{1}{\sqrt{x+4}}\right) = -\frac{1}{2(x+4)^\frac{3}{2}}\].

Find the Second Derivative of the Function

Continuing on, find the second derivative of \(f(x)\), using the chain rule again: \[f''(x) = \frac{d}{dx}\left(-\frac{1}{2(x+4)^\frac{3}{2}}\right) = \frac{3}{4(x+4)^\frac{5}{2}}\].

Evaluate the Second Derivative at the Given Point

Substitute \(x=0\) into the second derivative, yielding \[f''(0) = \frac{3}{4(0+4)^\frac{5}{2}} = \frac{3}{32}.\]

Key concepts

Chain Rule

When dealing with the differentiation of composite functions, the chain rule is an essential tool in calculus. It allows us to differentiate a function that is composed of one or more functions. Imagine the compound function as a set of interconnected gears, where the rotation of one affects the rotation of the others. Similarly, when a function, say, h(x), is the composition of two functions, f(g(x)), the chain rule helps us find h'(x) by multiplying the derivative of the outer function f'(g(x)) by the derivative of the inner function g'(x).

In the given exercise, the chain rule was applied twice: first, to determine the first derivative of the function, and then again to find the second derivative. A clear grasp of the chain rule is critical to progressing in calculus, as it comes up in various forms, whether you're dealing with simple functions or more complex expressions involving square roots, exponential functions, and others.

First Derivative

The first derivative of a function, represented as f'(x) or \frac{df}{dx}, is the backbone of differential calculus. It provides the rate at which the function's value changes concerning its input. Think of it as a snapshot of a car's speedometer, showing how fast the car is going at a particular moment. It helps us understand the behavior of functions and is a stepping stone towards finding higher-order derivatives.

In the provided example, finding the first derivative of the function \( f(x)=\frac{1}{\sqrt{x+4}} \) using the chain rule sets the stage for all subsequent steps, including the computation of the second derivative. The first derivative indicates how the slope of the function's graph changes as x varies, which leads to insights about the function's increasing or decreasing behavior and potential points of extrema.

Computer Algebra System

A Computer Algebra System (CAS) is a powerful software tool that can perform symbolic mathematics. As an advanced calculator, it doesn't just provide numerical solutions; it manipulates algebraic expressions, solving equations analytically, differentiating and integrating functions, factoring polynomials, and much more. In the context of our problem, a CAS could be used to verify the derivatives and the evaluated value of the second derivative at a point.

We encourage the student to both understand the foundational math and to utilize modern tools like a CAS to check work. By confirming the manually calculated second derivative of \( \frac{3}{4(x+4)^\frac{5}{2}} \) and its evaluation at the point \( x=0 \) with a CAS, students can gain confidence in their calculations and also familiarize themselves with the technology commonly used in advanced mathematics and engineering fields.

Function Evaluation

Function evaluation is akin to asking a 'what-if' question about a function. What if x equals a specific value? What happens to the function's output? When evaluating a function at a given point, we are essentially substituting the input with that value and simplifying to find the output. In our example, evaluating the second derivative at the point \( x=0 \) gives us specific information about the curvature of the function's graph at that point.

Evaluating the second derivative at a certain point helps us understand the concavity of the function at that point—whether the graph is curving upwards or downwards. With the evaluation showing \( f''(0) = \frac{3}{32} \) in our exercise, it indicates the graph of \( f(x) \) at \( x=0 \) is concave upwards, which can aid in graphing the function accurately or in understanding its geometric properties.