Question

Given \(y=\frac{2}{3} x^{3}-\frac{4}{x^{2}}+\frac{1}{2 x}-\sqrt{x}\), determine \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\)

Short answer

The second derivative is \( \frac{\mathrm{d}^{2}y}{\mathrm{d}x^2} = 4x - 24x^{-4} + x^{-3} + \frac{1}{4}x^{-3/2} \).

Step-by-step solution

Identify the Requirement

The exercise asks us to find the second derivative of the function with respect to x, i.e., \( \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} \). We will start by finding the first derivative and then differentiate it again.

Rewrite the Function for Differentiation

The given function is \( y=\frac{2}{3} x^{3}-\frac{4}{x^{2}}+\frac{1}{2 x}-\sqrt{x} \). Rewrite this as \( y = \frac{2}{3} x^3 - 4x^{-2} + \frac{1}{2}x^{-1} - x^{1/2} \) to make differentiation easier.

Differentiate to Find the First Derivative

Apply the power rule to each term: \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{3} \times 3x^2 - 4 \times (-2)x^{-3} + \frac{1}{2} \times (-1)x^{-2} - \frac{1}{2}x^{-1/2} \). Simplify the derivative: \( \frac{\mathrm{d}y}{\mathrm{d}x} = 2x^2 + 8x^{-3} - \frac{1}{2}x^{-2} - \frac{1}{2}x^{-1/2} \).

Differentiate Again to Find the Second Derivative

Differentiate the first derivative using the power rule: \( \frac{\mathrm{d}^{2}y}{\mathrm{d}x^2} = 4x - 24x^{-4} + x^{-3} + \frac{1}{4}x^{-3/2} \). Simplify the expression to find the second derivative: \( \frac{\mathrm{d}^{2}y}{\mathrm{d}x^2} = 4x - 24x^{-4} + x^{-3} + \frac{1}{4}x^{-3/2} \).

Key concepts

Understanding the Power Rule

The power rule is a fundamental concept in differentiation. If you have a function of the form \( f(x) = x^n \), where \( n \) is any real number, the derivative is calculated using the power rule: \( f'(x) = nx^{n-1} \). This rule holds for any real exponent, whether it's a positive integer, negative number, or a fraction.

By applying the power rule, differentiation becomes straightforward. Simply multiply the term by the exponent and reduce the exponent by one. Take for instance the function term \( \frac{2}{3}x^3 \), applying the power rule gives \( \frac{2}{3} \times 3x^{3-1} = 2x^2 \). Notice how each step just follows from the power rule straightforwardly.

When using the power rule, it’s crucial to first express each term in a polynomial form, just as we did by rewriting \( y \) in the exercise. This makes the differentiation process seamless.

Differentiation Explained

Differentiation is the process of finding a derivative, which represents the rate of change of a function with respect to its variable. For a function described as \( y = f(x) \), the first derivative \( \frac{dy}{dx} \) gives us the slope of the tangent line at any point \( x \).

The second derivative, \( \frac{d^2y}{dx^2} \), takes this a step further by providing information about the curvature or concavity of the original function. If \( \frac{d^2y}{dx^2} > 0 \), the graph is concave up, indicating a local minimum or a valley. Conversely, if \( \frac{d^2y}{dx^2} < 0 \), the graph is concave down, suggesting a local maximum or a hill.

In this exercise, differentiation required simplifying the function expression and then systematically applying the power rule to find both the first and second derivatives. Understanding these steps underscores the importance of derivatives in analyzing and understanding the behavior of functions.

Simplifying Expressions in Calculus

Simplifying expressions in calculus is a key step before applying rules like the power rule. The initial function provided, \( y = \frac{2}{3} x^3 - \frac{4}{x^2} + \frac{1}{2x} - \sqrt{x} \), poses challenges for direct application of the power rule due to terms in fractional and radical forms.

To simplify, each term is rewritten to align with polynomial expressions. For example, \( \frac{4}{x^2} \) becomes \( 4x^{-2} \), and \( \sqrt{x} \) turns into \( x^{1/2} \). This transformation ensures each term becomes compatible with the power rule, allowing differentiation to proceed smoothly.

By simplifying the function first, it's much easier to differentiate term by term without memorizing extra steps or special rules. Simplification transforms complex fractions and roots into exponents, making them ready for the easy application of the power rule.