# Derivative of the Difference of Functions

It is given that the derivative of a function that is the difference of two other functions, is equal to the difference of their derivatives. This can be proved by using derivative by definition or the 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
$\begin{gathered} \Rightarrow \Delta y = f\left( {x + \Delta x} \right) – g\left( {x + \Delta x} \right) – f\left( x \right) + g\left( x \right) \\ \Rightarrow \Delta y = f\left( {x + \Delta x} \right) – f\left( x \right) – g\left( {x + \Delta x} \right) + g\left( x \right) \\ \Rightarrow \Delta y = \left[ {f\left( {x + \Delta x} \right) – f\left( x \right)} \right] – \left[ {g\left( {x + \Delta x} \right) – g\left( x \right)} \right] \\ \end{gathered}$

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

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 difference of the two given functions is equal to the difference of their derivatives.

This difference rule can be expand to 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 $y = \left( {4{x^3}} \right) – \left( {6x + 5} \right)$

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

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

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