# Derivative of the Sum of Functions

It is given that the derivative of a function that is the sum of two other functions, is equal to the sum of their derivatives. This can be proved by using the derivative by definition or first principle method.

Consider a function of the form $y = f\left( x \right) + g\left( x \right)$.

First we take the increment or small change in the function.
$\begin{gathered} y + \Delta y = f\left( {x + \Delta x} \right) + g\left( {x + \Delta x} \right) \\ \Rightarrow \Delta y = f\left( {x + \Delta x} \right) + g\left( {x + \Delta x} \right) – y \\ \end{gathered}$

Putting the value of function $y = f\left( x \right) + g\left( x \right)$ in the above equation, we get
$\begin{gathered} \Rightarrow \Delta y = f\left( {x + \Delta x} \right) + g\left( {x + \Delta x} \right) – f\left( x \right) – g\left( x \right) \\ \Rightarrow \Delta y = f\left( {x + \Delta x} \right) – f\left( x \right) + g\left( {x + \Delta x} \right) – g\left( x \right) \\ \end{gathered}$

Dividing both sides by $\Delta x$, we get
$\begin{gathered}\frac{{\Delta y}}{{\Delta x}} = \frac{{f\left( {x + \Delta x} \right) – f\left( x \right) + g\left( {x + \Delta x} \right) – g\left( x \right)}}{{\Delta x}} \\ \frac{{\Delta y}}{{\Delta x}} = \frac{{f\left( {x + \Delta x} \right) – f\left( x \right)}}{{\Delta x}} + \frac{{g\left( {x + \Delta x} \right) – g\left( x \right)}}{{\Delta x}} \\ \end{gathered}$

Taking the limit of both sides as $\Delta x \to 0$, we have
$\begin{gathered} \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \left[ {\frac{{f\left( {x + \Delta x} \right) – f\left( x \right)}}{{\Delta x}} + \frac{{g\left( {x + \Delta x} \right) – g\left( x \right)}}{{\Delta x}}} \right] \\ \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) – f\left( x \right)}}{{\Delta x}} + \mathop {\lim }\limits_{\Delta x \to 0} \frac{{g\left( {x + \Delta x} \right) – g\left( x \right)}}{{\Delta x}} \\ \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) – f\left( x \right)}}{{\Delta x}} + \mathop {\lim }\limits_{\Delta x \to 0} \frac{{g\left( {x + \Delta x} \right) – g\left( x \right)}}{{\Delta x}} \\ \end{gathered}$

By the definition of derivative we have
$\frac{{dy}}{{dx}} = f’\left( x \right) + g’\left( x \right)$

This shows that the derivative of the sum of two given functions is equal to the sum of their derivatives.

This sum rule can be expanded more than two function as
$\frac{d}{{dx}}\left( {u + v + w + \cdots } \right) = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}} + \frac{{dw}}{{dx}} + \cdots$

Example: Find the derivative of $y = 4{x^3} + 6x + 5$

We have the given function as
$y = 4{x^3} + 6x + 5$

Differentiating with respect to variable $x$, we get
$\begin{gathered}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {4{x^3} + 6x + 5} \right) \\ \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {4{x^3}} \right) + \frac{d}{{dx}}\left( {6x + 5} \right) \\ \end{gathered}$

Now using the formula of derivatives, we have
$\begin{gathered} \frac{{dy}}{{dx}} = 12{x^2} + 6\left( 1 \right) + 0 \\ \frac{{dy}}{{dx}} = 12{x^2} + 6 \\ \end{gathered}$