Question

Use the table to find the following derivatives. $$\begin{array}{cccccc}x & 1 & 2 & 3 & 4 & 5 \\\\\hline f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\\g^{\prime}(x) & 2 & 4 & 3 & 1 & 5\end{array}$$ $$\left.\frac{d}{d x}(f(x)+g(x))\right|_{x=1}$$

Short answer

Answer: The value of the derivative of the sum of the functions f(x) and g(x) evaluated at x = 1 is 5.

Step-by-step solution

Understand the addition rule of derivatives

The addition rule for derivatives states that the derivative of the sum of two functions is the sum of their derivatives. In other words, $$\frac{d}{d x}(f(x) + g(x)) = \frac{d}{d x}f(x) + \frac{d}{d x}g(x)$$

Apply the rule to find the derivatives of f(x) + g(x)

Using the addition rule from Step 1, we know that we need to find the derivative of f(x) and g(x) individually and add them together. $$ \frac{d}{d x}(f(x) + g(x)) = f^{\prime}(x) + g^{\prime}(x) $$

Evaluate the derivatives at x = 1

Now, we will look up the values of the derivatives of f(x) and g(x) at x = 1 in the provided table. The table tells us that $$ f^{\prime}(1) = 3 \quad \text{and} \quad g^{\prime}(1) = 2 $$

Add the derivatives and find the final result

Finally, using the values of the derivatives found in Step 3, plug them into the equation from Step 2: $$ \frac{d}{d x}(f(x) + g(x))\bigg|_{x=1} = f^{\prime}(1) + g^{\prime}(1) = 3 + 2 = 5 $$ So, the value of the derivative of the sum of the functions f(x) and g(x) evaluated at x = 1 is 5.

Key concepts

Addition Rule of Derivatives

When studying calculus, one of the fundamental principles you'll encounter is the addition rule of derivatives. This rule is incredibly handy when you're dealing with functions that can be split into simpler parts. According to this rule, if you have two functions, say f(x) and g(x), and you want to take the derivative of their sum, you can simply take the derivatives of each function separately and add them together. In mathematical terms:
\[\frac{d}{d x}(f(x) + g(x)) = \frac{d}{d x}f(x) + \frac{d}{d x}g(x)\]
This concept becomes particularly valuable when functions are too complex to differentiate as a whole or when you have a list of derivative values available, as in the exercise. By breaking down the process step by step, it's clearer and easier to understand how each part contributes to the final result. For instance, if you know the derivatives of f(x) and g(x) at a specific point, you can simply add those known values to obtain the derivative of f(x) + g(x) at that point.

Derivative Evaluation

Derivative evaluation is the process of calculating the value of a derivative at a particular point. This is an essential technique in calculus, especially when analyzing the behavior of functions. Once you have the general expression for the derivative, evaluating it at a certain point can give you the slope of the tangent line to the original function at that point, indicating how fast the function is changing at that precise location.
For example, in the exercise, we see that you're asked to evaluate the derivative of the sum of two functions at x = 1. To do this, you first determine the individual derivatives of the functions at x = 1 from the given table, then sum them up using the addition rule previously explained. It's a straightforward process that illustrates how derivatives at specific points can provide meaningful information about a function's behavior. This direct use of derivative values at certain points further simplifies computations in practical applications, such as finding the rate of change in physics or the slope of a curve in geometry.

Calculus

Calculus is a branch of mathematics focusing on rates of change (differential calculus) and accumulation of quantities (integral calculus). It's a powerful tool that's used across various fields of science, economics, engineering, and beyond. In essence, calculus helps us understand and model the change and motion we see in the world around us.
Through concepts like derivatives and integrals, calculus provides the framework for solving problems that involve complex variables and their relationships. Returning to our exercise, by applying the principles of calculus, specifically the addition rule for derivatives, we can solve problems that initially seem complicated. The ability to break down these problems into smaller, more manageable parts is one of the reasons calculus is so widely applied. Whether examining the trajectory of a planet or optimizing a function for business efficiency, calculus forms the core of many critical analyses and decision-making processes.