Derivative of Sum of Functions

It is given that two functions are in form and take derivative of sum of these two functions which is equal to sum of their derivatives. This can be proving by using 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 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 definition of derivative we have

\frac{{dy}}{{dx}}  = f'\left( x \right) + g'\left( x \right)


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


Differentiation 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 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}

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