Let $f(x) = \sqrt{2x^2 + 2}$. Find $f^{\prime}(x$\].$f^{\prime}(x) =$

Let $f(x) = \sqrt{2x^2 + 2}$. Find $f^{\prime}(x)$

In this article, we will explore the concept of finding the derivative of a given function. The derivative of a function represents the rate of change of the function with respect to its input. In this case, we are given the function $f(x) = \sqrt{2x^2 + 2}$ and we need to find its derivative $f^{\prime}(x)$.

Before we dive into finding the derivative, let's understand the given function. The function $f(x) = \sqrt{2x^2 + 2}$ is a square root function that takes the input $x$ and returns the square root of $2x^2 + 2$. This function is a combination of a quadratic function and a square root function.

To find the derivative of the function $f(x) = \sqrt{2x^2 + 2}$, we can use the chain rule of differentiation. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of the function is given by $f^{\prime}(g(x)) \cdot g^{\prime}(x)$.

In this case, we can rewrite the function $f(x) = \sqrt{2x^2 + 2}$ as $f(x) = (2x^2 + 2)^{\frac{1}{2}}$. Now, we can apply the chain rule to find the derivative of the function.

Applying the Chain Rule

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

$$f^{\prime}(x) = \frac{d}{dx} (2x^2 + 2)^{\frac{1}{2}}$$

$$f^{\prime}(x) = \frac{1}{2} (2x^2 + 2)^{-\frac{1}{2}} \cdot \frac{d}{dx} (2x^2 + 2)$$

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

Simplifying the Derivative

Now, we can simplify the derivative by combining the terms:

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

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

In this article, we found the derivative of the function $f(x) = \sqrt{2x^2 + 2}$ using the chain rule of differentiation. The derivative of the function is given by $f^{\prime}(x) = \frac{2x}{\sqrt{2x^2 + 2}}$. This derivative represents the rate of change of the function with respect to its input.

The final answer is $\boxed{\frac{2x}{\sqrt{2x^2 + 2}}}$.
Q&A: Finding the Derivative of $f(x) = \sqrt{2x^2 + 2}$

In our previous article, we found the derivative of the function $f(x) = \sqrt{2x^2 + 2}$ using the chain rule of differentiation. In this article, we will answer some common questions related to finding the derivative of this function.

Q: What is the chain rule of differentiation?

A: The chain rule of differentiation is a fundamental concept in calculus that allows us to find the derivative of a composite function. A composite function is a function that is composed of two or more functions. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of the function is given by $f^{\prime}(g(x)) \cdot g^{\prime}(x)$.

Q: How do I apply the chain rule to find the derivative of $f(x) = \sqrt{2x^2 + 2}$?

A: To apply the chain rule, we need to identify the inner and outer functions of the composite function. In this case, the inner function is $2x^2 + 2$ and the outer function is the square root function. We can then find the derivative of the inner function and multiply it by the derivative of the outer function.

Q: What is the derivative of the inner function $2x^2 + 2$?

A: The derivative of the inner function $2x^2 + 2$ is given by:

$$\frac{d}{dx} (2x^2 + 2) = 4x$$

Q: What is the derivative of the outer function, the square root function?

A: The derivative of the outer function, the square root function, is given by:

$$\frac{d}{dx} (2x^2 + 2)^{\frac{1}{2}} = \frac{1}{2} (2x^2 + 2)^{-\frac{1}{2}}$$

Q: How do I combine the derivatives of the inner and outer functions to find the derivative of $f(x) = \sqrt{2x^2 + 2}$?

A: To combine the derivatives of the inner and outer functions, we multiply the derivative of the inner function by the derivative of the outer function:

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

Q: What is the final derivative of $f(x) = \sqrt{2x^2 + 2}$?

A: The final derivative of $f(x) = \sqrt{2x^2 + 2}$ is given by:

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

In this article, we answered some common questions related to finding the derivative of the function $f(x) = \sqrt{2x^2 + 2}$. We hope that this article has provided you with a better understanding of the chain rule of differentiation and how to apply it to find the derivative of a composite function.

The final answer is $\boxed{\frac{2x}{\sqrt{2x^2 + 2}}}$.