How do you differentiate $f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ using the product rule?

Hint: In this question we have to find the derivative of the term using the product rule. The product rule generally states that when the function is in the form of $g\left( x \right)\times h\left( x \right)$ the derivative of the function can be calculated as $g'\left( x \right)\times h\left( x \right)+h'\left( x \right)\times g\left( x \right)$. But in the above expression, we have three terms in multiplication which means that the function is in the form of $g\left( x \right)\times h\left( x \right)\times i\left( x \right)$. The formula to calculate the derivative of this function is $f'\left( x \right)=g'\left( x \right)\times h\left( x \right)\times i\left( x \right)+h'\left( x \right)\times g\left( x \right)\times i\left( x \right)+i'\left( x \right)\times g\left( x \right)\times h\left( x \right)$, which we will use and simplify to get the required solution.

Complete step-by-step solution:
We have the expression given to us as:
$f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$
We have to find the derivative of the term therefore; it can be written as:
$f'\left( x \right)=\dfrac{d}{dx}\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$
Now since the above expression is in the form of $g\left( x \right)\times h\left( x \right)\times i\left( x \right)$ , the derivative of the expression can be calculated by the formula $f'\left( x \right)=g'\left( x \right)\times h\left( x \right)\times i\left( x \right)+h'\left( x \right)\times g\left( x \right)\times i\left( x \right)+i'\left( x \right)\times g\left( x \right)\times h\left( x \right)$.
On using the formula, we get:
\[\Rightarrow f'\left( x \right)=\left( \dfrac{d}{dx}\left( x-1 \right) \right)\left( x-2 \right)\left( x-3 \right)+\left( \dfrac{d}{dx}\left( x-2 \right) \right)\left( x-1 \right)\left( x-3 \right)+\left( \dfrac{d}{dx}\left( x-3 \right) \right)\left( x-1 \right)\left( x-2 \right)\]
we know that $\dfrac{dx}{dx}=1$ and $\dfrac{d}{dx}k=0$ therefore, on using the formula, we get:
\[\Rightarrow f'\left( x \right)=\left( 1 \right)\left( x-2 \right)\left( x-3 \right)+\left( 1 \right)\left( x-1 \right)\left( x-3 \right)+\left( 1 \right)\left( x-1 \right)\left( x-2 \right)\]
Now on simplifying everything, we get:
\[\Rightarrow f'\left( x \right)=\left( {{x}^{2}}-5x+6 \right)+\left( {{x}^{2}}-4x+3 \right)+\left( {{x}^{2}}-3x+2 \right)\]
On opening the brackets and adding the similar terms, we get:
\[\Rightarrow f'\left( x \right)=3{{x}^{2}}-12x+11\], which is the required solution.

Note: In this question we have used the product rule on three terms. The general rule should be remembered for product rule that when there are terms in product the derivative can be calculated by taking the derivative of the first term and multiplying by the rest followed up by adding the derivative of the second term and multiplying the rest so forth up to the last term.
In this question it is acceptable to leave the solution before the line “simplifying everything”, but it is good practice to simplify in case substitutions have to be made.