The Product Rule of Derivatives

The product rule of derivatives is $\frac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f’\left( x \right)g\left( x \right) + f\left( x \right)g’\left( x \right)$. We can read this as the derivative of the product of two functions is equal to the derivative of the first function, multiplied with the second function as it is, plus the first function as it is, multiplied with the derivative of the second function. This product rule can be proved using the first principle or derivative by definition.

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
$\Rightarrow \Delta y = f\left( {x + \Delta x} \right)g\left( {x + \Delta x} \right) – f\left( x \right)g\left( x \right)$

Subtracting and adding $f\left( x \right)g\left( {x + \Delta x} \right)$ on the right hand side, we have
$\Rightarrow \Delta y = f\left( {x + \Delta x} \right)g\left( {x + \Delta x} \right) – f\left( x \right)g\left( {x + \Delta x} \right) + f\left( x \right)g\left( {x + \Delta x} \right) – f\left( x \right)g\left( x \right)$

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

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

NOTE: If we extended the product of three functions, then
$\Rightarrow \frac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)h\left( x \right)} \right] = f’\left( x \right)g\left( x \right)h\left( x \right) + f\left( x \right)g’\left( x \right)h\left( x \right) + f\left( x \right)g\left( x \right)h’\left( x \right)$

Example: Find the derivative of $y = \left( {2{x^2} + 5} \right)\left( {4x – 1} \right)$

We have the given function as
$y = \left( {2{x^2} + 5} \right)\left( {4x – 1} \right)$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {2{x^2} + 5} \right)\left( {4x – 1} \right)$

Now using the formula of the derivative of a square root, we have
$\begin{gathered}\frac{{dy}}{{dx}} = \left( {2{x^2} + 5} \right)\frac{d}{{dx}}\left( {4x – 1} \right) + \left( {4x – 1} \right)\frac{d}{{dx}}\left( {2{x^2} + 5} \right) \\ \frac{{dy}}{{dx}} = \left( {2{x^2} + 5} \right)\left( 4 \right) + \left( {4x – 1} \right)\left( {4x} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \left( {8{x^2} + 20} \right) + \left( {16{x^2} – 4x} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = 24{x^2} – 4x + 20 \\ \end{gathered}$