Find The Derivative Of The Function.$\[ Y = X^3 - \frac{x^2}{28} + 16x + 1 \\]$\[ \frac{d Y}{d X} = \square \\]

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Introduction

In calculus, the derivative of a function represents the rate of change of the function with respect to one of its variables. It is a fundamental concept in mathematics and has numerous applications in various fields, including physics, engineering, and economics. In this article, we will focus on finding the derivative of a polynomial function, specifically the function y=x3−x228+16x+1y = x^3 - \frac{x^2}{28} + 16x + 1.

What is a Derivative?

A derivative is a measure of how a function changes as its input changes. It is defined as the limit of the difference quotient of the function as the change in the input approaches zero. In other words, the derivative of a function f(x)f(x) is denoted by f′(x)f'(x) and represents the rate of change of the function with respect to xx.

Rules of Differentiation

There are several rules of differentiation that can be used to find the derivative of a function. These rules include:

  • Power Rule: If f(x)=xnf(x) = x^n, then f′(x)=nxn−1f'(x) = nx^{n-1}
  • Product Rule: If f(x)=u(x)v(x)f(x) = u(x)v(x), then f′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x)
  • Quotient Rule: If f(x)=u(x)v(x)f(x) = \frac{u(x)}{v(x)}, then f′(x)=u′(x)v(x)−u(x)v′(x)v(x)2f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}

Finding the Derivative of the Given Function

To find the derivative of the given function y=x3−x228+16x+1y = x^3 - \frac{x^2}{28} + 16x + 1, we will apply the power rule and the quotient rule.

Step 1: Differentiate the First Term

The first term of the function is x3x^3. Using the power rule, we can find its derivative as follows:

ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2

Step 2: Differentiate the Second Term

The second term of the function is −x228-\frac{x^2}{28}. To find its derivative, we will apply the quotient rule. Let u(x)=x2u(x) = x^2 and v(x)=28v(x) = 28. Then, u′(x)=2xu'(x) = 2x and v′(x)=0v'(x) = 0. Using the quotient rule, we get:

ddx(−x228)=−2x(28)−x2(0)282=−2x28=−x14\frac{d}{dx}\left(-\frac{x^2}{28}\right) = -\frac{2x(28) - x^2(0)}{28^2} = -\frac{2x}{28} = -\frac{x}{14}

Step 3: Differentiate the Third Term

The third term of the function is 16x16x. Using the power rule, we can find its derivative as follows:

ddx(16x)=16\frac{d}{dx}(16x) = 16

Step 4: Differentiate the Fourth Term

The fourth term of the function is 11. Since the derivative of a constant is zero, we have:

ddx(1)=0\frac{d}{dx}(1) = 0

Combining the Derivatives

Now that we have found the derivatives of each term, we can combine them to find the derivative of the entire function:

dydx=3x2−x14+16\frac{d y}{d x} = 3x^2 - \frac{x}{14} + 16

Conclusion

In this article, we have found the derivative of the function y=x3−x228+16x+1y = x^3 - \frac{x^2}{28} + 16x + 1 using the power rule and the quotient rule. The derivative of the function is given by dydx=3x2−x14+16\frac{d y}{d x} = 3x^2 - \frac{x}{14} + 16. This result can be used to analyze the behavior of the function and to solve optimization problems.

Applications of Derivatives

Derivatives have numerous applications in various fields, including:

  • Physics: Derivatives are used to describe the motion of objects and to calculate forces and energies.
  • Engineering: Derivatives are used to design and optimize systems, such as electrical circuits and mechanical systems.
  • Economics: Derivatives are used to model economic systems and to make predictions about future economic trends.

Final Thoughts

In conclusion, finding the derivative of a function is a fundamental concept in calculus that has numerous applications in various fields. By applying the power rule and the quotient rule, we can find the derivative of a polynomial function and use it to analyze the behavior of the function and to solve optimization problems.

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Introduction

In our previous article, we discussed how to find the derivative of a polynomial function using the power rule and the quotient rule. In this article, we will answer some frequently asked questions about derivatives and provide additional examples to help you understand the concept better.

Q: What is the derivative of a constant function?

A: The derivative of a constant function is zero. This is because the derivative represents the rate of change of the function with respect to its input, and a constant function does not change as its input changes.

Q: How do I find the derivative of a function with multiple terms?

A: To find the derivative of a function with multiple terms, you can apply the power rule and the quotient rule separately to each term and then combine the results. For example, if you have a function like y=x3+2x2−3x+1y = x^3 + 2x^2 - 3x + 1, you can find the derivative of each term separately and then combine them to get the final derivative.

Q: What is the difference between the power rule and the quotient rule?

A: The power rule is used to find the derivative of a function that is raised to a power, such as x3x^3. The quotient rule is used to find the derivative of a function that is the ratio of two functions, such as x2x+1\frac{x^2}{x+1}.

Q: Can I use the derivative to find the original function?

A: Yes, you can use the derivative to find the original function. This is known as the process of integration. However, integration is a more complex process than differentiation, and it requires a different set of rules and techniques.

Q: How do I use the derivative to solve optimization problems?

A: To use the derivative to solve optimization problems, you need to find the critical points of the function, which are the points where the derivative is equal to zero or undefined. You can then use the second derivative test to determine whether the critical point is a maximum, minimum, or saddle point.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine whether a critical point is a maximum, minimum, or saddle point. It involves finding the second derivative of the function and evaluating it at the critical point. If the second derivative is positive, the critical point is a minimum. If the second derivative is negative, the critical point is a maximum. If the second derivative is zero, the test is inconclusive.

Q: Can I use the derivative to model real-world phenomena?

A: Yes, you can use the derivative to model real-world phenomena. For example, you can use the derivative to model the motion of an object, the growth of a population, or the spread of a disease.

Q: What are some common applications of derivatives?

A: Some common applications of derivatives include:

  • Physics: Derivatives are used to describe the motion of objects and to calculate forces and energies.
  • Engineering: Derivatives are used to design and optimize systems, such as electrical circuits and mechanical systems.
  • Economics: Derivatives are used to model economic systems and to make predictions about future economic trends.
  • Biology: Derivatives are used to model the growth of populations and the spread of diseases.

Conclusion

In conclusion, derivatives are a powerful tool for modeling and analyzing real-world phenomena. By understanding the concept of derivatives and how to apply them, you can solve a wide range of problems in physics, engineering, economics, and other fields. We hope this article has helped you to better understand the concept of derivatives and how to use them to solve problems.