The **Sum Rule,** also known as the * Addition Rule,* is a fundamental principle in differentiation used to find the derivative of a sum of two or more functions. This rule simplifies the process of differentiating such functions and is widely used in various applications of calculus, including optimization problems, physics, and engineering.

## Sum Rule Formula and Formal Definition

The Sum Rule formula for differentiation is as follows:

$$\frac{\text{d}}{\text{d}x}(f(x)+g(x))=\frac{\text{d}}{\text{d}x}f(x)+\frac{\text{d}}{\text{d}x}g(x)$$**Definition:**
For differentiable functions $f(x)$ and $g(x)$, the derivative of the sum of these functions with respect to $x$ is given by:

where $\frac{\text{d}}{\text{d}x}f(x)$ is the derivative of $f(x)$ and $\frac{\text{d}}{\text{d}x}g(x)$ is the derivative of $g(x)$.

The Sum Rule can be extended to the sum of any finite number of differentiable functions:

$$\frac{\text{d}}{\text{d}x}({f}_{1}(x)+{f}_{2}(x)+\cdots +{f}_{n}(x))=\frac{\text{d}}{\text{d}x}{f}_{1}(x)+\frac{\text{d}}{\text{d}x}{f}_{2}(x)+\cdots +\frac{\text{d}}{\text{d}x}{f}_{n}(x)$$## Steps to Apply the Sum Rule

**Identify the functions:**Determine the individual functions $f(x)$, $g(x)$, or any other functions in the given sum.**Find the derivatives of each function:**Differentiate each function in the sum with respect to its variable using the appropriate differentiation rules.**Add the derivatives:**Add the derivatives of the individual functions obtained in step 2.**Simplify the result:**If necessary, simplify the final expression obtained from step 3.

## Example: Applying the Sum Rule

Let’s consider the function $h(x)={x}^{3}+\mathrm{sin}(x)$.

The functions in the sum are $f(x)={x}^{3}$ and $g(x)=\mathrm{sin}(x)$.

The derivative of ${x}^{3}$ is $3{x}^{2}$ (using the Power Rule), and the derivative of $\mathrm{sin}(x)$ is $\mathrm{cos}(x)$.

Adding the derivatives:

$${h}^{\prime}(x)=\frac{\text{d}}{\text{d}x}({x}^{3}+\mathrm{sin}(x))=\frac{\text{d}}{\text{d}x}{x}^{3}+\frac{\text{d}}{\text{d}x}\mathrm{sin}(x)=3{x}^{2}+\mathrm{cos}(x)$$The result is already simplified, so we have:

$${h}^{\prime}(x)=3{x}^{2}+\mathrm{cos}(x)$$

Thus, the derivative of $h(x)={x}^{3}+\mathrm{sin}(x)$ is ${h}^{\prime}(x)=3{x}^{2}+\mathrm{cos}(x)$.

## Sum Rule Proof

To prove the Sum Rule for differentiation, we will use the definition of the derivative (i.e., the first principle of differentiation) and the limit properties. Let $h(x)=f(x)+g(x)$, where $f(x)$ and $g(x)$ are differentiable functions.

Proof:

$$\begin{array}{cc}\hfill {h}^{\prime}(x)& =\frac{\text{d}}{\text{d}x}(f(x)+g(x))\hfill \\ \hfill [\mathrm{4ex}]& ={lim}_{h\to 0}\frac{(f(x+h)+g(x+h))-(f(x)+g(x))}{h}\hfill \\ \hfill [\mathrm{4ex}]& ={lim}_{h\to 0}[\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}]\hfill \\ \hfill [\mathrm{4ex}]& ={lim}_{h\to 0}\frac{f(x+h)-f(x)}{h}+{lim}_{h\to 0}\frac{g(x+h)-g(x)}{h}\hfill \\ \hfill [\mathrm{4ex}]& ={f}^{\prime}(x)+{g}^{\prime}(x)\hfill \end{array}$$Therefore, we have proved that $\frac{\text{d}}{\text{d}x}(f(x)+g(x))=\frac{\text{d}}{\text{d}x}f(x)+\frac{\text{d}}{\text{d}x}g(x)$.

## Detailed Explanation of the Proof Steps

$$h(x)=f(x)+g(x)$$**Start with the definition of $h(x)$:**$h(x)$ is defined as the sum of $f(x)$ and $g(x)$:

$${h}^{\prime}(x)=\frac{\text{d}}{\text{d}x}(f(x)+g(x))$$**Definition of the derivative:**The derivative of $h(x)$ with respect to $x$ is given by:

$${h}^{\prime}(x)={lim}_{h\to 0}\frac{(f(x+h)+g(x+h))-(f(x)+g(x))}{h}$$**Apply the limit definition of the derivative:**Using the limit definition, the derivative can be written as:

$${h}^{\prime}(x)={lim}_{h\to 0}\frac{(f(x+h)+g(x+h))-f(x)-g(x)}{h}$$**Distribute the terms in the numerator:**Rewrite the expression inside the limit by distributing the terms:

$${h}^{\prime}(x)={lim}_{h\to 0}[\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}]$$**Separate the fractions:**Split the fraction into two separate fractions:

$${h}^{\prime}(x)={lim}_{h\to 0}\frac{f(x+h)-f(x)}{h}+{lim}_{h\to 0}\frac{g(x+h)-g(x)}{h}$$**Use the sum of limits property:**Apply the property that the limit of a sum is the sum of the limits:

$${h}^{\prime}(x)={f}^{\prime}(x)+{g}^{\prime}(x)$$**Recognize the individual derivatives:**Each term inside the limits represents the derivative of the respective functions:

Thus, we have proved that:

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