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Problem #1: Let $f(x) = \frac{1}{x^2 + 1}$ and $g(x) = \frac{x}{x^2 + 1}$. Find the derivative of the function $h(x) = f(x) + g(x)$.

Step 1: Recall the Sum Rule of Differentiation

The sum rule of differentiation states that if we have two functions $f(x)$ and $g(x)$, then the derivative of their sum is given by:

$$\frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$$

Step 2: Find the Derivative of $f(x)$

To find the derivative of $f(x) = \frac{1}{x^2 + 1}$, we can use the quotient rule of differentiation. However, in this case, we can simplify the function by recognizing that it is the derivative of the arctangent function. Specifically, we have:

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

Using the chain rule, we can find the derivative of $f(x)$ as:

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

Step 3: Find the Derivative of $g(x)$

To find the derivative of $g(x) = \frac{x}{x^2 + 1}$, we can use the quotient rule of differentiation. The quotient rule states that if we have two functions $u(x)$ and $v(x)$, then the derivative of their quotient is given by:

$$\frac{d}{dx} \left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \frac{d}{dx} u(x) - u(x) \frac{d}{dx} v(x)}{(v(x))^2}$$

In this case, we have $u(x) = x$ and $v(x) = x^2 + 1$. Using the quotient rule, we can find the derivative of $g(x)$ as:

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

Simplifying the expression, we get:

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

Step 4: Find the Derivative of $h(x)$

Using the sum rule of differentiation, we can find the derivative of $h(x) = f(x) + g(x)$ as:

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

Simplifying the expression, we get:

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

The final answer is: $\boxed{\frac{2 - x^2}{(x^2 + 1)^2}}$

Problem #2: Let $f(x) = \frac{1}{x^2 + 1}$ and $g(x) = \frac{x}{x^2 + 1}$. Find the derivative of the function $h(x) = f(x) + g(x)$.

Q&A: Derivatives and Functions

Q: What is the derivative of the function $f(x) = \frac{1}{x^2 + 1}$?

A: The derivative of the function $f(x) = \frac{1}{x^2 + 1}$ is given by:

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

Q: What is the derivative of the function $g(x) = \frac{x}{x^2 + 1}$?

A: The derivative of the function $g(x) = \frac{x}{x^2 + 1}$ is given by:

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

Q: How do we find the derivative of the function $h(x) = f(x) + g(x)$?

A: To find the derivative of the function $h(x) = f(x) + g(x)$, we can use the sum rule of differentiation. The sum rule states that if we have two functions $f(x)$ and $g(x)$, then the derivative of their sum is given by:

$$\frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$$

Q: What is the derivative of the function $h(x) = f(x) + g(x)$?

A: The derivative of the function $h(x) = f(x) + g(x)$ is given by:

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

Q: What is the final answer to the problem?

A: The final answer to the problem is:

$$\boxed{\frac{2 - x^2}{(x^2 + 1)^2}}$$

Additional Questions and Answers

Q: What is the quotient rule of differentiation?

A: The quotient rule of differentiation states that if we have two functions $u(x)$ and $v(x)$, then the derivative of their quotient is given by:

$$\frac{d}{dx} \left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \frac{d}{dx} u(x) - u(x) \frac{d}{dx} v(x)}{(v(x))^2}$$

Q: What is the chain rule of differentiation?

A: The chain rule of differentiation states that if we have a composite function $f(g(x))$, then the derivative of the function is given by:

$$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$$

Q: What is the product rule of differentiation?

A: The product rule of differentiation states that if we have two functions $f(x)$ and $g(x)$, then the derivative of their product is given by:

$$\frac{d}{dx} (f(x) \cdot g(x)) = f(x) \frac{d}{dx} g(x) + g(x) \frac{d}{dx} f(x)$$

Conclusion

In this article, we have discussed the problem of finding the derivative of the function $h(x) = f(x) + g(x)$, where $f(x) = \frac{1}{x^2 + 1}$ and $g(x) = \frac{x}{x^2 + 1}$. We have used the sum rule of differentiation, the quotient rule of differentiation, and the chain rule of differentiation to find the derivative of the function. We have also answered additional questions and provided explanations for the rules of differentiation.