Question

What is the derivative of \[y = (1 - x)(2 - x)...(n - x)\,at\,x = 1\] ?

Answer 1

Hint: Here in the given question we need to solve the given expression, by using property of differentiation, in which we can differentiate multiple brackets together, which says, that first differentiate the first bracket then write all left bracket in product form, then put add sign and differentiate the second bracket and write the rest of bracket in product form, and so on.
Formulae Used:
\[
   \Rightarrow \dfrac{d}{{dx}}(u)(v)(w) = u'(v)(w) + (u)v'(w) + (u)(v)w’ \\
   \Rightarrow here\,u' = \dfrac{{du}}{{dx}} \\
 \]
\[ \Rightarrow (n - 1)! = 1 \times 2 \times ... \times (n - 1)\]

Complete step by step answer:
Here in the above question, to solve the expression, we need to use the property of differentiation, of multiple brackets together, on solving we get:
We have:
 \[ \Rightarrow y = (1 - x)(2 - x)...(n - x)\]
Now using formulae:
\[ \Rightarrow \dfrac{d}{{dx}}(u)(v)(w) = u'(v)(w) + (u)v'(w) + (u)(v)w’\]
We get:
\[
   \Rightarrow \dfrac{{dy}}{{dx}} = (1 - x)'(2 - x)...(n - x) + (1 - x)(2 - x)'...(n - x) + ... \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - 1(2 - x)...(n - x) + 0 + ...0 \\
 \]
Here only the first bracket differentiation will give the positive value, rest will give zero value for “x=1”, since the first bracket is “(1-x)” and it will become zero for “x=1”, on solving we get:
\[
   \Rightarrow \dfrac{{dy}}{{dx}} = - 1(2 - x)...(n - x) + 0 + ...0 \\
   \Rightarrow for\,x = 1 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - 1(2 - 1)...(n - 1) = - 1 \times (1) \times (2)...(n - 1) \\
 \]
Now here we obtain a product of “n-1”, terms, and we know that with the help of factorial we can write the answer her, on solving we get:
We have:
\[ \Rightarrow (n - 1)! = 1 \times 2 \times ... \times (n - 1)\]
Now putting this in our solution we get:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - 1 \times (n - 1)! = - (n - 1)!\]
Here we obtain the derivative of the required expression.

Note: Here in the above question, we find the derivation in one step above the final solution, but to write the simplified solution we use here factorial, we know factorial represents the product of the given number to the unity, in descending manner.