If $f(x)=\frac{4 X^2+4 X+7}{\sqrt{x}}$, Then:1. Find The Derivative, $f^{\prime}(x$\].2. Evaluate The Derivative At $x = 5$, $f^{\prime}(5$\].

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 explore the derivative of a given function and evaluate it at a specific point.

The Given Function

The given function is:

$$f(x)=\frac{4 x^2+4 x+7}{\sqrt{x}}$$

This function is a rational function, where the numerator is a quadratic expression and the denominator is a square root of $x$. To find the derivative of this function, we will use the quotient rule and the chain rule.

Finding the Derivative

To find the derivative of $f(x)$, we will use the quotient rule, which states that if $f(x) = \frac{g(x)}{h(x)}$, then $f^{\prime}(x) = \frac{h(x)g^{\prime}(x) - g(x)h^{\prime}(x)}{[h(x)]^2}$.

In this case, $g(x) = 4x^2 + 4x + 7$ and $h(x) = \sqrt{x}$. We will first find the derivatives of $g(x)$ and $h(x)$.

Derivative of $g(x)$

The derivative of $g(x)$ is:

$$g^{\prime}(x) = \frac{d}{dx}(4x^2 + 4x + 7) = 8x + 4$$

Derivative of $h(x)$

The derivative of $h(x)$ is:

$$h^{\prime}(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$$

Applying the Quotient Rule

Now, we will apply the quotient rule to find the derivative of $f(x)$:

$$f^{\prime}(x) = \frac{h(x)g^{\prime}(x) - g(x)h^{\prime}(x)}{[h(x)]^2}$$

Substituting the expressions for $g(x)$, $h(x)$, $g^{\prime}(x)$, and $h^{\prime}(x)$, we get:

$$f^{\prime}(x) = \frac{\sqrt{x}(8x + 4) - (4x^2 + 4x + 7)\frac{1}{2\sqrt{x}}}{x}$$

Simplifying the expression, we get:

$$f^{\prime}(x) = \frac{8x^{3/2} + 4x^{1/2} - 2x - \frac{7}{2\sqrt{x}}}{x}$$

Evaluating the Derivative at $x = 5$

To evaluate the derivative at $x = 5$, we will substitute $x = 5$ into the expression for $f^{\prime}(x)$:

$$f^{\prime}(5) = \frac{8(5)^{3/2} + 4(5)^{1/2} - 2(5) - \frac{7}{2\sqrt{5}}}{5}$$

Simplifying the expression, we get:

$$f^{\prime}(5) = \frac{8(11.1803) + 4(2.2361) - 10 - \frac{7}{2(2.2361)}}{5}$$

$$f^{\prime}(5) = \frac{89.361 + 8.9444 - 10 - 1.5625}{5}$$

$$f^{\prime}(5) = \frac{86.7409}{5}$$

$$f^{\prime}(5) = 17.34818$$

Conclusion

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Q&A: Derivatives and Function Evaluation

In this article, we will answer some common questions related to derivatives and function evaluation.

Q: What is the derivative of a function?

A: 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.

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

A: To find the derivative of a function, you can use various rules, including the power rule, the product rule, the quotient rule, and the chain rule. The power rule states that if $f(x) = x^n$, then $f^{\prime}(x) = nx^{n-1}$. The product rule states that if $f(x) = g(x)h(x)$, then $f^{\prime}(x) = g^{\prime}(x)h(x) + g(x)h^{\prime}(x)$. The quotient rule states that if $f(x) = \frac{g(x)}{h(x)}$, then $f^{\prime}(x) = \frac{h(x)g^{\prime}(x) - g(x)h^{\prime}(x)}{[h(x)]^2}$. The chain rule states that if $f(x) = g(h(x))$, then $f^{\prime}(x) = g^{\prime}(h(x))h^{\prime}(x)$.

Q: What is the difference between the derivative and the integral of a function?

A: The derivative of a function represents the rate of change of the function with respect to its input, while the integral of a function represents the accumulation of the function over a given interval. The derivative is used to find the rate of change of a function, while the integral is used to find the area under a curve or the accumulation of a function over a given interval.

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 can substitute the value of the point into the expression for the derivative. For example, if you want to evaluate the derivative of $f(x) = x^2$ at $x = 3$, you would substitute $x = 3$ into the expression for the derivative, which is $f^{\prime}(x) = 2x$. This would give you $f^{\prime}(3) = 2(3) = 6$.

Q: What are some common applications of derivatives and function evaluation?

A: Derivatives and function evaluation have numerous applications in various fields, including physics, engineering, economics, and computer science. Some common applications include:

• Finding the rate of change of a function
• Finding the maximum or minimum value of a function
• Finding the area under a curve or the accumulation of a function over a given interval
• Modeling real-world phenomena, such as population growth or chemical reactions
• Optimizing functions, such as finding the maximum or minimum value of a function subject to certain constraints

Q: What are some common mistakes to avoid when working with derivatives and function evaluation?

A: Some common mistakes to avoid when working with derivatives and function evaluation include:

• Forgetting to apply the chain rule or the product rule when differentiating a function
• Making errors when substituting values into expressions for derivatives
• Failing to check the domain of a function before evaluating its derivative
• Not considering the context of the problem when evaluating a derivative or integral

Conclusion

In this article, we have answered some common questions related to derivatives and function evaluation. We have discussed the definition of a derivative, the rules for finding derivatives, and some common applications of derivatives and function evaluation. We have also highlighted some common mistakes to avoid when working with derivatives and function evaluation. By understanding these concepts and avoiding common mistakes, you can become proficient in working with derivatives and function evaluation.