Let \[$ F(x) = -4x^8 \ln X \$\].Find The Derivative:\[$ F^{\prime}(x) = \$\]Evaluate The Derivative At \[$ X = E^2 \$\]:\[$ F^{\prime}(e^2) = \$\]

Feb 27, 2025 by ADMIN 147 views

Let's Dive into Calculus: Finding the Derivative of a Complex Function

In this article, we will explore the concept of finding the derivative of a complex function, specifically the function ${ f(x) = -4x^8 \ln x }$ . We will use the power rule and the product rule of differentiation to find the derivative of this function. Additionally, we will evaluate the derivative at a specific point, ${ x = e^2 }$ .

Understanding the Function

Before we dive into finding the derivative, let's take a closer look at the function ${ f(x) = -4x^8 \ln x }$ . This function is a product of two functions: ${ -4x^8 }$ and ${ \ln x }$ . The first function is a polynomial function, while the second function is a natural logarithm function.

Finding the Derivative

To find the derivative of the function ${ f(x) = -4x^8 \ln x }$ , we will use the product rule of differentiation. The product rule states that if we have a function of the form ${ f(x) = u(x)v(x) }$ , then the derivative of the function is given by ${ f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x) }$ .

In this case, we have ${ u(x) = -4x^8 }$ and ${ v(x) = \ln x }$ . To find the derivative of ${ u(x) }$ , we will use the power rule of differentiation, which states that if we have a function of the form ${ f(x) = x^n }$ , then the derivative of the function is given by ${ f^{\prime}(x) = nx^{n-1} }$ .

Applying the power rule to ${ u(x) = -4x^8 }$ , we get ${ u^{\prime}(x) = -32x^7 }$ . To find the derivative of ${ v(x) = \ln x }$ , we will use the fact that the derivative of the natural logarithm function is given by ${ \frac{d}{dx} \ln x = \frac{1}{x} }$ .

Therefore, we have ${ v^{\prime}(x) = \frac{1}{x} }$ . Now, we can use the product rule to find the derivative of the function ${ f(x) = -4x^8 \ln x }$ .

The Derivative of the Function

Using the product rule, we get:

${ f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x) }$

${ f^{\prime}(x) = (-32x^7)(\ln x) + (-4x^8)\left(\frac{1}{x}\right) }$

${ f^{\prime}(x) = -32x^7 \ln x - 4x^7 }$

Evaluating the Derivative

Now that we have found the derivative of the function, we can evaluate it at a specific point, ${ x = e^2 }$ . To do this, we will substitute ${ x = e^2 }$ into the derivative of the function.

${ f^{\prime}(e^2) = -32(e^2)^7 \ln (e^2) - 4(e^2)^7 }$

${ f^{\prime}(e^2) = -32e^{14} \ln e^2 - 4e^{14} }$

${ f^{\prime}(e^2) = -32e^{14} (2) - 4e^{14} }$

${ f^{\prime}(e^2) = -64e^{14} - 4e^{14} }$

${ f^{\prime}(e^2) = -68e^{14} }$

Conclusion

In this article, we have found the derivative of the complex function ${ f(x) = -4x^8 \ln x }$ using the product rule of differentiation. We have also evaluated the derivative at a specific point, ${ x = e^2 }$ . The derivative of the function is given by ${ f^{\prime}(x) = -32x^7 \ln x - 4x^7 }$ , and the value of the derivative at ${ x = e^2 }$ is ${ f^{\prime}(e^2) = -68e^{14} }$ .

Key Takeaways

The product rule of differentiation states that if we have a function of the form ${ f(x) = u(x)v(x) }$ , then the derivative of the function is given by ${ f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x) }$ .
The power rule of differentiation states that if we have a function of the form ${ f(x) = x^n }$ , then the derivative of the function is given by ${ f^{\prime}(x) = nx^{n-1} }$ .
The derivative of the natural logarithm function is given by ${ \frac{d}{dx} \ln x = \frac{1}{x} }$ .

Further Reading

If you are interested in learning more about calculus and differentiation, I recommend checking out the following resources:

Khan Academy: Calculus
MIT OpenCourseWare: Calculus
Wolfram MathWorld: Calculus

References

Stewart, J. (2016). Calculus: Early Transcendentals. Cengage Learning.
Anton, H. (2016). Calculus: A New Horizon. John Wiley & Sons.
Edwards, C. H. (2016). Calculus: Early Transcendentals. Pearson Education.
Q&A: Calculus and Differentiation

In this article, we will answer some common questions about calculus and differentiation. Whether you are a student looking for help with your homework or a professional looking to brush up on your skills, this article is for you.

Q: What is calculus?

A: Calculus is a branch of mathematics that deals with the study of continuous change. It is a fundamental tool for understanding many phenomena in the natural world, from the motion of objects to the growth of populations.

Q: What is differentiation?

A: Differentiation is a key concept in calculus that involves finding the rate of change of a function with respect to one of its variables. It is a way of measuring how fast a function changes as its input changes.

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

A: There are several ways to find the derivative of a function, depending on the type of function and the rules of differentiation that apply. Some common rules include the power rule, the product rule, and the quotient rule.

Q: What is the power rule of differentiation?

A: The power rule of differentiation states that if we have a function of the form ${ f(x) = x^n }$ , then the derivative of the function is given by ${ f^{\prime}(x) = nx^{n-1} }$ .

Q: What is the product rule of differentiation?

A: The product rule of differentiation states that if we have a function of the form ${ f(x) = u(x)v(x) }$ , then the derivative of the function is given by ${ f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x) }$ .

Q: What is the quotient rule of differentiation?

A: The quotient rule of differentiation states that if we have a function of the form ${ f(x) = \frac{u(x)}{v(x)} }$ , then the derivative of the function is given by ${ f^{\prime}(x) = \frac{u^{\prime}(x)v(x) - u(x)v^{\prime}(x)}{(v(x))^2} }$ .

Q: How do I evaluate the derivative of a function at a specific point?

A: To evaluate the derivative of a function at a specific point, you simply substitute the value of the point into the derivative of the function.

Q: What is the chain rule of differentiation?

A: The chain rule of differentiation states that if we have a function of the form ${ f(x) = g(h(x)) }$ , then the derivative of the function is given by ${ f^{\prime}(x) = g^{\prime}(h(x)) \cdot h^{\prime}(x) }$ .

Q: What is the fundamental theorem of calculus?

A: The fundamental theorem of calculus states that differentiation and integration are inverse processes. This means that if we have a function and its derivative, we can use the fundamental theorem of calculus to find the original function.

Q: What are some common applications of calculus?

A: Calculus has many applications in a wide range of fields, including physics, engineering, economics, and computer science. Some common applications include:

Modeling population growth and decay
Describing the motion of objects
Finding the maximum and minimum values of functions
Calculating the area and volume of complex shapes
Modeling the behavior of electrical circuits

Q: What are some common mistakes to avoid when working with calculus?

A: Some common mistakes to avoid when working with calculus include:

Failing to check units and dimensions
Failing to simplify expressions
Failing to use the correct rules of differentiation
Failing to evaluate limits correctly
Failing to check for extraneous solutions