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Find the first and second derivatives. \begin{equation} y=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\frac{x}{4} \end{equation}

Short Answer

Expert verified
First derivative: \( y' = x^2 + x + \frac{1}{4} \). Second derivative: \( y'' = 2x + 1 \).

Step by step solution

01

Identify the Function

The given function is \( y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4} \). Our task is to find the first and second derivatives of this function with respect to \( x \).
02

Find the First Derivative

To find the first derivative \( y' \), apply the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \). For each term in our function, apply the rule as follows:1. \( \frac{d}{dx} \left( \frac{x^3}{3} \right) = 3 \cdot \frac{1}{3} x^{3-1} = x^2 \) 2. \( \frac{d}{dx} \left( \frac{x^2}{2} \right) = 2 \cdot \frac{1}{2} x^{2-1} = x \) 3. \( \frac{d}{dx} \left( \frac{x}{4} \right) = \frac{1}{4} \cdot 1x^{1-1} = \frac{1}{4} \) Therefore, the first derivative is \( y' = x^2 + x + \frac{1}{4} \).
03

Find the Second Derivative

To find the second derivative \( y'' \), differentiate the first derivative \( y' = x^2 + x + \frac{1}{4} \) again:1. \( \frac{d}{dx}(x^2) = 2x \) 2. \( \frac{d}{dx}(x) = 1 \) 3. \( \frac{d}{dx} \left( \frac{1}{4} \right) = 0 \) Therefore, the second derivative is \( y'' = 2x + 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a fundamental concept in calculus, specifically in differentiation. This rule helps us determine the derivative of a function in the form of a power of a variable. In simpler terms, when you have a function that is raised to some power, the power rule provides an easy method to find the derivative.Let's say you have a function \( x^n \), where \( n \) is any real number. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). The process involves:
  • Taking the exponent and multiplying it by the coefficient of the term.
  • Reducing the exponent by one to reflect the new power of \( x \).
This rule is especially handy when dealing with polynomial functions, just like in the original exercise. It streamlines the differentiation process and is one of the first rules students learn when starting calculus.
First Derivative
The first derivative of a function provides critical information about the rate of change of the function's value concerning the independent variable, often \( x \). It's essentially the slope of the function or the velocity at which it is changing.To find the first derivative, we apply differentiation techniques, such as the power rule. For example, in the exercise provided:1. Start with the function \( y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4} \).2. Apply the power rule to each term individually: - The derivative of \( \frac{x^3}{3} \) is \( x^2 \). - The derivative of \( \frac{x^2}{2} \) is \( x \). - The derivative of \( \frac{x}{4} \) is \( \frac{1}{4} \).Ultimately, the first derivative is expressed as \( y' = x^2 + x + \frac{1}{4} \).This derivative tells us how the function \( y \) changes as \( x \) changes.
Second Derivative
The second derivative provides insights into the acceleration of a function's rate of change. It tells us how the rate at which a function grows or shrinks is changing over time. This can be used to discern the concavity of the function:
  • If the second derivative is positive, the function is concave up, resembling a "U" shape.
  • If negative, the function is concave down, showing an inverted "U" shape.
Continuing from the first derivative \( y' = x^2 + x + \frac{1}{4} \):1. Differentiate once more to find the second derivative: - \( \, \frac{d}{dx}(x^2) = 2x \, \) - \( \, \frac{d}{dx}(x) = 1 \, \) - \( \, \frac{d}{dx}(\frac{1}{4}) = 0 \, \)Combining these, the second derivative is \( y'' = 2x + 1 \). This tells us how the slope \( y' \) itself is changing.
Differentiation
Differentiation is a core concept in calculus, focusing on analyzing how functions change. It involves finding the derivative, which measures a function's sensitivity to changes in its inputs. In simple terms: - Differentiation helps find the slope or gradient of a function at any given point. - It allows us to study a function's behavior in a localized manner. The process involves applying rules such as the power rule, product rule, or quotient rule, depending on the nature of the function. In the context of the original exercise, differentiation is used to calculate both the first and second derivatives. Differentiation is an essential tool in a wide array of applications, from physics to economics, and underscores much of the theoretical underpinnings of calculus.
Calculus
Calculus is a branch of mathematics that studies continuous change. It's made up of two main areas: differentiation and integration. Differentiation is concerned with the concept of a derivative, while integration deals with the concept of an integral. In practical terms, calculus provides:
  • Methods for solving problems involving changing quantities.
  • Techniques for calculating areas, volumes, and other quantities.
The specific exercise illustrates the differentiation aspect of calculus, showcasing how to apply the power rule to find derivatives. Calculus equips us with a comprehensive toolkit for modeling and solving real-world scenarios, from motion and growth phenomena to optimizing functions in various scientific disciplines.

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