Derivative of the Difference of Functions

It is given that two functions are in form and take derivative of difference of these two functions which is equal to difference 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) \\ \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 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 difference of two given functions is equal to the difference of their derivatives.

This difference 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 = \left( {4{x^3}} \right) - \left( {6x + 5} \right)

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}} \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}