Question

Evaluating a Second Derivative In Exercises \(91-94\) , evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result. $$ f(x)=\frac{1}{\sqrt{x+4}},\left(0, \frac{1}{2}\right) $$

Short answer

The second derivative of the function at the point \( (0, \frac{1}{2}) \) is \( \frac{3}{32} \)

Step-by-step solution

Calculate the first derivative

Derive the function \(f(x)= \frac{1}{\sqrt{x+4}} \). Here we have to apply the chain rule for differentiation (which states that the derivative of the composition of two functions is the derivative of the first times the composition of the second). The chain rule is applicable here where the outer function is \(f(u) = \frac{1}{u}\) and the inner function is \(u = \sqrt{x+4}\). Therefore, the first derivative is \(f'(x) = - \frac{{1}}{{2\sqrt{x+4}^3}}\).

Calculate the second derivative

To find the second derivative, we need to differentiate the first derivative \(f'(x) = - \frac{{1}}{{2\sqrt{x+4}^3}} \) that we just found. When differentiating \(f'(x)\) again to get the second derivative, use the chain rule again. The second derivative is \(f''(x) = \frac{{3}}{{4(x+4)^{5/2}}}\).

Evaluate the second derivative at the given point

The given point is (0, 1/2). In this problem, we only need the x-coordinate to evaluate the second derivative. So for \(x = 0\), substitute \(0\) into the second derivative equation. This gives us \(f''(0) = \frac{{3}}{{4(0+4)^{5/2}}} = \frac{3}{32}\).

Key concepts

Chain Rule for Differentiation

The chain rule is a staple in calculus, particularly when dealing with the differentiation of composite functions. It enables you to find the derivative of a function that is composed of two or more functions, essentially 'nested' within each other. Imagine a function within a function, much like a Russian doll scenario.

Here's how it works: If you have a function denoted as \( f(x) = g(h(x)) \), then the derivative of \( f \) with respect to \( x \), denoted as \( f'(x) \), is \( g'(h(x)) \times h'(x) \). In simpler terms, you differentiate the outer function (as if the inner function were a simple variable), and then multiply it by the derivative of the inner function.

Applying this to our exercise, the outer function is \( f(u) = 1/u \), and the inner function is \( u = \/{x + 4}\). So we differentiate \( 1/u \) with respect to \( u \) to get \( -1/u^2 \), then multiply by the derivative of \( \/{x + 4}\) which is \( 1/(2\sqrt{x+4}) \). This gives us the first derivative, namely \( f'(x) = -\frac{1}{2\sqrt{x+4}^3} \), following the chain rule for differentiation.

Derivatives in Calculus

Derivatives can be thought of as the mathematical expression for the rate at which something changes. In calculus, finding a derivative means determining this rate of change at any point on a function. It is fundamental to understanding many aspects of change, motion, and growth across various scientific fields.

When we speak of a 'second derivative', as in our exercise, we're looking at the rate at which the rate of change is changing. It's like analyzing the acceleration of a car when you already know its speed. The second derivative can reveal information about concavity and points of inflection on a graph, which are crucial for understanding the behavior of functions. In the exercise, the second derivative \( f''(x) = \frac{3}{4(x+4)^{\frac{5}{2}}} \) tells us how the slope of the tangent line to the function's graph is changing as \( x \) changes.

Computer Algebra System

A Computer Algebra System (CAS) is like a deluxe toolbox for mathematicians, engineers, scientists, and students. It's a software program capable of carrying out symbolic mathematical manipulations. This includes a variety of tasks such as solving equations, finding derivatives and integrals, simplifying complex expressions, and even verifying solutions.

Imagine you're working through a tough calculus problem and you want to ensure your manual calculations are correct. Here's where a CAS becomes invaluable. It offers a way to double-check work, explore mathematical concepts, and perform computations that would be exceedingly tedious or prone to error if done by hand. For instance, in our exercise, after we manually calculated the second derivative, we could use a CAS to verify that the computation of \( f''(0) = \frac{3}{32} \) is indeed correct, avoiding any miscalculations and enhancing our confidence in the result.