Product Rule of Derivatives, Definition and Examples

Product Rule of Derivatives:If $f$ and $g$ are differentiable at $x$ , then $f_g$ is also differentiable at $x$ and

$\lbrack f(x)g(x)\rbrack'=f(x)g'(x)+f'(x)g(x)$ .

$i.e,\;\;\;\;\;\frac d{dx}\lbrack f(x)g(x)\rbrack=f(x)\frac d{dx}g(x)+g(x)\frac d{dx}f(x)$ .

$\phi(x\;)=f\left(x\right)\;g\left(x\right)\;...........................(1)$

$\phi(x+\delta x\;)=f\left(x+\delta x\right)\;g\left(x+\delta x\right)\;\;........(2)$ .

Subtracting  .

$\phi(x+\delta x\;)-\phi(x)=f\left(x+\delta x\right)\;g\left(x+\delta x\right)\;-\;f\left(x\right)\;g\left(x\right)$ .

Subtracting and Adding .

$\phi(x+\delta x\;)-\phi(x)=f\left(x+\delta x\right)\;g\left(x+\delta x\right)\;-\;f\left(x\right)\;g\left(x+\delta x\right)+f\left(x\right)g\left(x+\delta x\right)-f\left(x\right)\;g\left(x\right)$ .

$\phi(x+\delta x\;)-\phi(x)=\left[f\left(x+\delta x\right)-f\left(x\right)\right]\;g\left(x+\delta x\right)\;+\;f\left(x\right)\left[g\left(x+\delta x\right)-\;g\left(x\right)\right]$ .

$\begin{array}{l}\frac{\phi(x+\delta x)-\phi(x)}{\delta x}=\left[\frac{\left[f\left(x+\delta x\right)-f\left(x\right)\right]}{\delta x}\right]g\left(x+\delta x\right)+f\left(x\right)\left[\frac{\left[g\left(x+\delta x\right)-g\left(x\right)\right]}{\delta x}\right]\end{array}$ .

                     Taking.

$\begin{array}{l}\underset{\delta x\rightarrow o}{lim}\left[\frac{\left[\phi\left(x+\delta x\right)-\phi\left(x\right)\right]}{\delta x}\right]=\underset{\delta x\rightarrow o}{lim}\left[\frac{\left[f\left(x+\delta x\right)-f\left(x\right)\right]}{\delta x}\right]g\left(x+\delta x\right)+f\left(x\right)\left[\frac{\left[g\left(x+\delta x\right)-g\left(x\right)\right]}{\delta x}\right]\end{array}$ .

$\begin{array}{l}\underset{\delta x\rightarrow o}{lim}\left[\frac{\left[\phi\left(x+\delta x\right)-\phi\left(x\right)\right]}{\delta x}\right]=\underset{\delta x\rightarrow o}{lim}\frac{\left[f\left(x+\delta x\right)-f\left(x\right)\right]}{\delta x}\underset{\delta x\rightarrow o}{lim}g\left(x+\delta x\right)+\underset{\delta x\rightarrow o}{lim}f\left(x\right)\underset{\delta x\rightarrow o}{lim}\frac{\left[g\left(x+\delta x\right)-g\left(x\right)\right]}{\delta x}\end{array}$ .

$\phi'(x\;)=f'\left(x\right)\;g\left(x\right)+f\left(x\right)\;g'\left(x\right)\;\;\;\;\;\;\;\;\;;\underset{\delta x\rightarrow0}{lim}g\left(x+\delta x\right)=g(x)$

$\frac d{dx}\;\left[f\left(x\right)\;g\left(x\right)\;\right]=\frac d{dx}\;\left[f\left(x\right)\right]\;g\left(x\right)+\frac d{dx}\;\left[g\left(x\right)\right]\;f\left(x\right)$ .

Example 1  {Product rule of derivatives}

Let $f(x)=3x$ and $g(x)=5x^3$

Now we use product Rule

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=\frac d{dx}\left[3x.5x^3\right]$

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=\left(5x^3\right)\frac d{dx}\left(3x\right)+3x\frac d{dx}\left(5x^3\right)$

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=\left(5x^3\right)\left(3\right)\frac d{dx}\left(x\right)-3x\left(5\right)\frac d{dx}\left(x^3\right)$

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=\left(5x^3\right)\left(3\right)\left(1\right)-3x\left(5\right)\left(3x^2\right)$

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=15x^3-30x^3$

$\boxed{\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=-15x^3}$

Example 2    [Product rule of derivatives]

Let $f(x)=3x^2$ and $g(x)=7x$

Now we use product rule

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=\frac d{dx}\left(3x^2.7x\right)$

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=7x\frac d{dx}\left(3x^2\right)+3x^2\frac d{dx}\left(7x\right)$

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=7x\left(3\right)\frac d{dx}\left(x^2\right)+3x^2\left(7\right)\frac d{dx}\left(x\right)$

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=7x\left(3\right)\left(2x\right)+3x^2\left(7\right)\left(1\right)$

$\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=42x^2+21x^2$

$\boxed{\frac d{dx}\left[f\left(x\right).g\left(x\right)\right]=63x^2}$

you can also see quotient rule

Leave a Comment Cancel Reply