Product Rule of Derivatives

Product rule of derivative 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). In words we can read as derivative of product of two functions is equal to derivative of first function second function as it is plus first function as it is derivative of second function. This product rule can be proving using 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 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 product of three function, 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)


Differentiation 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 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}

Comments

comments