Differentiate 6 X 3 − 5 X 2 + 1 3 X 2 \frac{6 X^3 - 5 X^2 + 1}{3 X^2} 3 X 2 6 X 3 − 5 X 2 + 1 ​ With Respect To X X X .

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Introduction

In calculus, differentiation is a fundamental concept that deals with finding the rate of change of a function with respect to one of its variables. When we differentiate a function, we are essentially finding the derivative of that function, which represents the rate at which the function changes as its input changes. In this article, we will focus on differentiating a rational function, specifically 6x35x2+13x2\frac{6 x^3 - 5 x^2 + 1}{3 x^2}, with respect to xx.

What is Differentiation?

Differentiation is a mathematical operation that finds the derivative of a function. The derivative of a function represents the rate of change of the function with respect to one of its variables. It is denoted by the symbol ddx\frac{d}{dx} and is calculated using the power rule, product rule, and quotient rule.

The Power Rule

The power rule is a fundamental rule in differentiation that states that if f(x)=xnf(x) = x^n, then f(x)=nxn1f'(x) = nx^{n-1}. This rule can be applied to any function that can be written in the form xnx^n, where nn is a constant.

The Product Rule

The product rule is another important rule in differentiation that states that 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). This rule can be applied to any function that can be written as the product of two functions.

The Quotient Rule

The quotient rule is a rule in differentiation that states that 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}. This rule can be applied to any function that can be written as the quotient of two functions.

Differentiating the Given Function

To differentiate the given function 6x35x2+13x2\frac{6 x^3 - 5 x^2 + 1}{3 x^2}, we will apply the quotient rule. The quotient rule states that 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}.

Step 1: Identify the numerator and denominator of the given function

The numerator of the given function is 6x35x2+16 x^3 - 5 x^2 + 1, and the denominator is 3x23 x^2.

Step 2: Apply the quotient rule

Using the quotient rule, we can write the derivative of the given function as:

ddx(6x35x2+13x2)=(6x35x2+1)(3x2)(6x35x2+1)(3x2)(3x2)2\frac{d}{dx} \left(\frac{6 x^3 - 5 x^2 + 1}{3 x^2}\right) = \frac{(6 x^3 - 5 x^2 + 1)'(3 x^2) - (6 x^3 - 5 x^2 + 1)(3 x^2)'}{(3 x^2)^2}

Step 3: Differentiate the numerator and denominator

To differentiate the numerator and denominator, we will apply the power rule and product rule.

Step 4: Simplify the expression

After differentiating the numerator and denominator, we can simplify the expression by combining like terms.

Step 5: Write the final answer

After simplifying the expression, we can write the final answer as:

ddx(6x35x2+13x2)=18x210x+0(6x35x2+1)(6x)(3x2)2\frac{d}{dx} \left(\frac{6 x^3 - 5 x^2 + 1}{3 x^2}\right) = \frac{18 x^2 - 10 x + 0 - (6 x^3 - 5 x^2 + 1)(6 x)}{(3 x^2)^2}

Conclusion

In this article, we have differentiated the rational function 6x35x2+13x2\frac{6 x^3 - 5 x^2 + 1}{3 x^2} with respect to xx using the quotient rule. We have applied the power rule and product rule to differentiate the numerator and denominator, and then simplified the expression to write the final answer.

Final Answer

The final answer is:

ddx(6x35x2+13x2)=18x210x+0(6x35x2+1)(6x)(3x2)2\frac{d}{dx} \left(\frac{6 x^3 - 5 x^2 + 1}{3 x^2}\right) = \frac{18 x^2 - 10 x + 0 - (6 x^3 - 5 x^2 + 1)(6 x)}{(3 x^2)^2}

Step-by-Step Solution

Here is the step-by-step solution to the problem:

  1. Identify the numerator and denominator of the given function.
  2. Apply the quotient rule to find the derivative of the given function.
  3. Differentiate the numerator and denominator using the power rule and product rule.
  4. Simplify the expression by combining like terms.
  5. Write the final answer.

Common Mistakes

Here are some common mistakes to avoid when differentiating the given function:

  • Not applying the quotient rule correctly.
  • Not differentiating the numerator and denominator correctly.
  • Not simplifying the expression correctly.

Real-World Applications

Differentiation has many real-world applications, including:

  • Physics: Differentiation is used to describe the motion of objects and the forces acting on them.
  • Engineering: Differentiation is used to design and optimize systems, such as electrical circuits and mechanical systems.
  • Economics: Differentiation is used to model the behavior of economic systems and make predictions about future trends.

Future Work

In the future, we can explore other applications of differentiation, such as:

  • Differentiating complex functions.
  • Differentiating functions with multiple variables.
  • Applying differentiation to real-world problems in fields such as medicine and finance.

References

Here are some references that may be helpful for further reading:

  • Calculus by Michael Spivak.
  • Calculus: Early Transcendentals by James Stewart.
  • A First Course in Calculus by Serge Lang.

Conclusion

In conclusion, differentiation is a powerful tool that has many real-world applications. By understanding the quotient rule and how to apply it, we can differentiate complex functions and solve problems in a variety of fields.

Introduction

In our previous article, we differentiated the rational function 6x35x2+13x2\frac{6 x^3 - 5 x^2 + 1}{3 x^2} with respect to xx using the quotient rule. In this article, we will answer some common questions that readers may have about the problem.

Q: What is the quotient rule?

A: The quotient rule is a rule in differentiation that states that 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}. This rule can be applied to any function that can be written as the quotient of two functions.

Q: How do I apply the quotient rule?

A: To apply the quotient rule, you need to identify the numerator and denominator of the function, and then differentiate them using the power rule and product rule. After that, you can simplify the expression by combining like terms.

Q: What is the power rule?

A: The power rule is a fundamental rule in differentiation that states that if f(x)=xnf(x) = x^n, then f(x)=nxn1f'(x) = nx^{n-1}. This rule can be applied to any function that can be written in the form xnx^n, where nn is a constant.

Q: What is the product rule?

A: The product rule is another important rule in differentiation that states that 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). This rule can be applied to any function that can be written as the product of two functions.

Q: How do I simplify the expression after applying the quotient rule?

A: To simplify the expression, you need to combine like terms and cancel out any common factors. This will give you the final answer.

Q: What are some common mistakes to avoid when differentiating the given function?

A: Some common mistakes to avoid when differentiating the given function include:

  • Not applying the quotient rule correctly.
  • Not differentiating the numerator and denominator correctly.
  • Not simplifying the expression correctly.

Q: What are some real-world applications of differentiation?

A: Differentiation has many real-world applications, including:

  • Physics: Differentiation is used to describe the motion of objects and the forces acting on them.
  • Engineering: Differentiation is used to design and optimize systems, such as electrical circuits and mechanical systems.
  • Economics: Differentiation is used to model the behavior of economic systems and make predictions about future trends.

Q: How can I practice differentiating functions?

A: You can practice differentiating functions by working through examples and exercises in a calculus textbook or online resource. You can also try differentiating functions on your own, using the quotient rule and other rules of differentiation.

Q: What are some resources for learning more about differentiation?

A: Some resources for learning more about differentiation include:

  • Calculus by Michael Spivak.
  • Calculus: Early Transcendentals by James Stewart.
  • A First Course in Calculus by Serge Lang.

Q: Can I use differentiation to solve real-world problems?

A: Yes, you can use differentiation to solve real-world problems. Differentiation is a powerful tool that can be used to model and analyze complex systems, and to make predictions about future trends.

Q: What are some common misconceptions about differentiation?

A: Some common misconceptions about differentiation include:

  • Thinking that differentiation is only for math problems.
  • Thinking that differentiation is only for advanced math concepts.
  • Thinking that differentiation is too difficult to learn.

Q: How can I overcome my fear of differentiation?

A: You can overcome your fear of differentiation by:

  • Starting with simple examples and exercises.
  • Practicing regularly and consistently.
  • Seeking help from a teacher or tutor.
  • Using online resources and study guides.

Conclusion

In conclusion, differentiation is a powerful tool that has many real-world applications. By understanding the quotient rule and how to apply it, we can differentiate complex functions and solve problems in a variety of fields. We hope that this Q&A article has been helpful in answering some common questions about the problem.