Differentiate The Function $G(x)=13 \sqrt{x} \sec X$.Answer:$G^{\prime}(x)=$

Differentiating the Function $G(x)=13 \sqrt{x} \sec x$

In this article, we will explore the process of differentiating the given function $G(x)=13 \sqrt{x} \sec x$. Differentiation is a fundamental concept in calculus that deals with the rate of change of a function with respect to its input. It is a crucial tool in various fields such as physics, engineering, and economics. The function $G(x)$ is a product of two functions, $\sqrt{x}$ and $\sec x$, and we will use the product rule of differentiation to find its derivative.

The product rule of differentiation states that if we have a function of the form $f(x) = u(x)v(x)$, where $u(x)$ and $v(x)$ are both functions of $x$, then the derivative of $f(x)$ with respect to $x$ is given by:

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

This rule can be extended to the case where we have a product of more than two functions.

Applying the Product Rule to $G(x)$

To find the derivative of $G(x)$, we will apply the product rule of differentiation. We have:

$$G(x) = 13 \sqrt{x} \sec x$$

We can rewrite this as:

$$G(x) = 13x^{\frac{1}{2}} \sec x$$

Now, we can identify the two functions $u(x)$ and $v(x)$ as:

$$u(x) = 13x^{\frac{1}{2}}$$

$$v(x) = \sec x$$

We can now find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

$$u^{\prime}(x) = \frac{13}{2}x^{-\frac{1}{2}}$$

$$v^{\prime}(x) = \sec x \tan x$$

Now, we can apply the product rule of differentiation to find the derivative of $G(x)$:

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

$$G^{\prime}(x) = \frac{13}{2}x^{-\frac{1}{2}} \sec x + 13x^{\frac{1}{2}} \sec x \tan x$$

We can simplify the derivative of $G(x)$ by combining the two terms:

$$G^{\prime}(x) = \frac{13}{2}x^{-\frac{1}{2}} \sec x + 13x^{\frac{1}{2}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2}x^{-\frac{1}{2}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2}x^{-\frac{1}{2}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

G^{\prime}(<br/> **Q&A: Differentiating the Function $G(x)=13 \sqrt{x} \sec x$** **Q: What is the derivative of the function $G(x)=13 \sqrt{x} \sec x$?** ============================================================= A: The derivative of the function $G(x)=13 \sqrt{x} \sec x$ is given by: $G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x

$$

A: To apply the product rule of differentiation, you need to identify the two functions $u(x)$ and $v(x)$ that make up the product $G(x)$. In this case, we have:

$$u(x) = 13x^{\frac{1}{2}}$$

$$v(x) = \sec x$$

You then need to find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

$$u^{\prime}(x) = \frac{13}{2}x^{-\frac{1}{2}}$$

$$v^{\prime}(x) = \sec x \tan x$$

Finally, you can apply the product rule of differentiation to find the derivative of $G(x)$:

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

$$G^{\prime}(x) = \frac{13}{2}x^{-\frac{1}{2}} \sec x + 13x^{\frac{1}{2}} \sec x \tan x$$

$$

A: Yes, you can simplify the derivative of $G(x)$ further by combining the two terms:

$$G^{\prime}(x) = \frac{13}{2}x^{-\frac{1}{2}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$G^{\prime}(x) = \frac{13}{2} \frac{1}{\sqrt{x}} \sec x + \frac{13}{\sqrt{x}} \sec x \tan x$$

$$

A: The derivative of $G(x)$ represents the rate of change of the function $G(x)$ with respect to its input $x$. It is a fundamental concept in calculus that deals with the rate of change of a function with respect to its input.

$$

A: The derivative of $G(x)$ can be used in various real-world applications, such as:

• Finding the maximum or minimum of a function
• Determining the rate of change of a function
• Solving optimization problems
• Modeling real-world phenomena

A: Yes, you can use the product rule of differentiation to find the derivative of other functions that are products of two or more functions. The product rule of differentiation states that if we have a function of the form $f(x) = u(x)v(x)$, where $u(x)$ and $v(x)$ are both functions of $x$, then the derivative of $f(x)$ with respect to $x$ is given by:

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

This rule can be extended to the case where we have a product of more than two functions.