Find { F^ \prime}(x) $}$ For The Given Function. Then Find { F^{\prime}(2), F^{\prime}(0) $}$, And { F^{\prime}(-2) $}$.Given { F(x) = 14 \sqrt{x $}$Find: { F^{\prime}(x) = $}$
Introduction
In calculus, the derivative of a function represents the rate of change of the function with respect to its input. 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 given function and then evaluating it at specific points.
The Given Function
The given function is { f(x) = 14 \sqrt{x} $}$. This function represents a square root function with a constant multiplier of 14.
Finding the Derivative
To find the derivative of the given function, we will use the power rule of differentiation. The power rule states that if { f(x) = x^n $}$, then { f^{\prime}(x) = nx^{n-1} $}$.
In this case, we have { f(x) = 14 \sqrt{x} $}$, which can be rewritten as { f(x) = 14x^{1/2} $}$. Applying the power rule, we get:
{ f^{\prime}(x) = 14 \cdot \frac{1}{2} x^{1/2 - 1} $}$
Simplifying the expression, we get:
{ f^{\prime}(x) = 7x^{-1/2} $}$
Evaluating the Derivative at Specific Points
Now that we have found the derivative of the given function, we can evaluate it at specific points. We are asked to find { f^{\prime}(2), f^{\prime}(0) $}$, and { f^{\prime}(-2) $}$.
To evaluate the derivative at these points, we simply substitute the values of x into the derivative expression:
{ f^{\prime}(2) = 7(2)^{-1/2} $}$
{ f^{\prime}(0) = 7(0)^{-1/2} $}$
{ f^{\prime}(-2) = 7(-2)^{-1/2} $}$
Evaluating these expressions, we get:
{ f^{\prime}(2) = 7 \cdot \frac{1}{\sqrt{2}} $}$
{ f^{\prime}(0) = 7 \cdot \frac{1}{\sqrt{0}} $}$
{ f^{\prime}(-2) = 7 \cdot \frac{1}{-\sqrt{2}} $}$
Simplifying these expressions, we get:
{ f^{\prime}(2) = \frac{7}{\sqrt{2}} $}$
{ f^{\prime}(0) = \text{undefined} $}$
{ f^{\prime}(-2) = -\frac{7}{\sqrt{2}} $}$
Conclusion
In this article, we found the derivative of the given function { f(x) = 14 \sqrt{x} $}$ using the power rule of differentiation. We then evaluated the derivative at specific points, including { f^{\prime}(2), f^{\prime}(0) $}$, and { f^{\prime}(-2) $}$. The results show that the derivative is undefined at x = 0, while it is positive at x = 2 and negative at x = -2.
Derivative Notation
In this article, we used the following notation to represent the derivative:
{ f^{\prime}(x) $}$
This notation represents the derivative of the function f(x) with respect to x.
Common Derivatives
Here are some common derivatives that you may find useful:
- { \frac{d}{dx} x^n = nx^{n-1} $}$
- { \frac{d}{dx} e^x = e^x $}$
- { \frac{d}{dx} \ln x = \frac{1}{x} $}$
- { \frac{d}{dx} \sin x = \cos x $}$
- { \frac{d}{dx} \cos x = -\sin x $}$
Practice Problems
Here are some practice problems to help you reinforce your understanding of derivatives:
- Find the derivative of the function { f(x) = 3x^2 + 2x - 5 $}$.
- Evaluate the derivative of the function { f(x) = x^3 - 2x^2 + x - 1 $}$ at x = 2.
- Find the derivative of the function { f(x) = \sin x + \cos x $}$.
- Evaluate the derivative of the function { f(x) = e^x + \ln x $}$ at x = 1.
- Find the derivative of the function { f(x) = x^2 \sin x $}$.
Solutions
Here are the solutions to the practice problems:
- { f^{\prime}(x) = 6x + 2 $}$
- { f^{\prime}(2) = 2(2)^3 - 4(2)^2 + 1 = 16 - 16 + 1 = 1 $}$
- { f^{\prime}(x) = \cos x - \sin x $}$
- { f^{\prime}(1) = e^1 + \frac{1}{1} = e + 1 $}$
- { f^{\prime}(x) = 2x \sin x + x^2 \cos x $}$
Conclusion
Q: What is a derivative?
A: A derivative is a measure of how a function changes as its input changes. It represents the rate of change of the function with respect to its input.
Q: How do I find the derivative of a function?
A: There are several rules for finding the derivative of a function, including the power rule, the product rule, and the quotient rule. The power rule states that if { f(x) = x^n $}$, then { f^{\prime}(x) = nx^{n-1} $}$. The product rule states that if { f(x) = u(x)v(x) $}$, then { f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x) $}$. The quotient rule states that if { f(x) = \frac{u(x)}{v(x)} $}$, then { f^{\prime}(x) = \frac{u^{\prime}(x)v(x) - u(x)v{\prime}(x)}{v(x)2} $}$.
Q: What is the difference between a derivative and a differential?
A: A derivative represents the rate of change of a function with respect to its input, while a differential represents the change in the function's output due to a small change in its input.
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 input into the derivative expression.
Q: What is the significance of the derivative in real-world applications?
A: The derivative has numerous applications in various fields, including physics, engineering, and economics. It is used to model the behavior of physical systems, optimize functions, and make predictions about future outcomes.
Q: Can you provide some examples of derivatives in real-world applications?
A: Yes, here are some examples of derivatives in real-world applications:
- Physics: The derivative is used to describe the motion of objects, including velocity and acceleration.
- Engineering: The derivative is used to design and optimize systems, including control systems and signal processing.
- Economics: The derivative is used to model the behavior of economic systems, including supply and demand curves.
Q: How do I use the derivative to optimize a function?
A: To use the derivative to optimize a function, you first find the derivative of the function and then set it equal to zero. This will give you the critical points of the function, which are the points where the function may have a maximum or minimum value.
Q: What is the relationship between the derivative and the second derivative?
A: The second derivative is the derivative of the first derivative. It is used to determine the concavity of a function and to find the inflection points.
Q: Can you provide some examples of the second derivative in real-world applications?
A: Yes, here are some examples of the second derivative in real-world applications:
- Physics: The second derivative is used to describe the acceleration of an object, including the rate of change of velocity.
- Engineering: The second derivative is used to design and optimize systems, including control systems and signal processing.
- Economics: The second derivative is used to model the behavior of economic systems, including the rate of change of economic growth.
Q: How do I use the derivative to model the behavior of a physical system?
A: To use the derivative to model the behavior of a physical system, you first find the derivative of the system's equations of motion and then use it to describe the system's behavior over time.
Q: What is the relationship between the derivative and the integral?
A: The derivative and the integral are inverse operations. The derivative represents the rate of change of a function with respect to its input, while the integral represents the accumulation of a function over a given interval.
Q: Can you provide some examples of the relationship between the derivative and the integral?
A: Yes, here are some examples of the relationship between the derivative and the integral:
- Physics: The derivative is used to describe the motion of objects, while the integral is used to describe the accumulation of energy over time.
- Engineering: The derivative is used to design and optimize systems, while the integral is used to describe the accumulation of signal over time.
- Economics: The derivative is used to model the behavior of economic systems, while the integral is used to describe the accumulation of economic growth over time.
Conclusion
In this article, we have provided a comprehensive guide to derivatives, including their definition, rules for finding them, and applications in real-world scenarios. We have also provided some examples of derivatives in real-world applications and discussed the relationship between the derivative and the second derivative.