Question

The derivative of $y = \left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)$ at $x = 1$ is equal to:
(A) $0$
(B) $\left( { - 1} \right)\left( {n - 1} \right)!$
(C) $n! - 1$
(D) ${\left( { - 1} \right)^{n - 1}}\left( {n - 1} \right)!$
(E) ${\left( { - 1} \right)^n}\left( {n - 1} \right)!$

Answer 1

In the given problem, we are required to differentiate $y = \left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)$ with respect to x and then substitute the value of x as one in the expression. To differentiate the given function with ease, we will take natural logarithm on both sides of the equation. Then, we will differentiate both sides with respect to x using the chain rule of differentiation.

Complete answer:
So, we have the function $y = \left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)$.
Taking natural logarithm on both sides of equation, we get,
$\Rightarrow \ln y = \ln \left[ {\left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)} \right]$
Now, we know the property of logarithm $\log xy = \log x + \log y$. So, we get,
$\Rightarrow \ln y = \ln \left( {1 - x} \right) + \ln \left( {2 - x} \right) + .... + \ln \left( {n - x} \right)$
Now, we know the derivative of natural logarithm is $\left( {\dfrac{1}{x}} \right)$. Differentiating both sides with respect to x, we get,
$\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{1}{{\left( {1 - x} \right)}} \times \dfrac{{dy}}{{dx}}\left( {1 - x} \right) + \dfrac{1}{{\left( {2 - x} \right)}} \times \dfrac{{dy}}{{dx}}\left( {2 - x} \right) + .... + \dfrac{1}{{\left( {n - x} \right)}} \times \dfrac{{dy}}{{dx}}\left( {n - x} \right)$
Using the chain rule of differentiation $f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right)g'\left( x \right)$, we get,
$\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{1}{{\left( {1 - x} \right)}} \times \left( { - 1} \right) + \dfrac{1}{{\left( {2 - x} \right)}} \times \left( { - 1} \right) + .... + \dfrac{1}{{\left( {n - x} \right)}} \times \left( { - 1} \right)$
Taking $\left( { - 1} \right)$ common from all the terms, we get,
$\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left[ {\dfrac{1}{{\left( {1 - x} \right)}} + \dfrac{1}{{\left( {2 - x} \right)}} + .... + \dfrac{1}{{\left( {n - x} \right)}}} \right]$
$\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left[ {\dfrac{1}{{\left( {1 - x} \right)}} + \dfrac{1}{{\left( {2 - x} \right)}} + .... + \dfrac{1}{{\left( {n - x} \right)}}} \right]$
Now, we find the value of $\dfrac{{dy}}{{dx}}$ by shifting terms in the equation and substituting the values of y in terms of x.
$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = y\left( { - 1} \right)\left[ {\dfrac{1}{{\left( {1 - x} \right)}} + \dfrac{1}{{\left( {2 - x} \right)}} + .... + \dfrac{1}{{\left( {n - x} \right)}}} \right]$
$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)\left[ {\dfrac{1}{{\left( {1 - x} \right)}} + \dfrac{1}{{\left( {2 - x} \right)}} + .... + \dfrac{1}{{\left( {n - x} \right)}}} \right]$
Opening the brackets and simplifying the expression, we get,
$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left[ {\dfrac{{\left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)}}{{\left( {1 - x} \right)}} + \dfrac{{\left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)}}{{\left( {2 - x} \right)}} + .... + \dfrac{{\left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)}}{{\left( {n - x} \right)}}} \right]$Cancelling the common factors in all the terms, we get,
$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left[ {\left( {2 - x} \right)\left( {3 - x} \right)....\left( {n - x} \right) + \left( {1 - x} \right)\left( {3 - x} \right)....\left( {n - x} \right) + .... + \left( {1 - x} \right)\left( {2 - x} \right)....\left( {\left( {n - 1} \right) - x} \right)} \right]$ $\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left[ {\left( {2 - x} \right)\left( {3 - x} \right)....\left( {n - x} \right) + \left( {1 - x} \right)\left( {3 - x} \right)....\left( {n - x} \right) + .... + \left( {1 - x} \right)\left( {2 - x} \right)....\left( {\left( {n - 1} \right) - x} \right)} \right]$
Now, putting the value of x as one, we get,
$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left[ {\left( {2 - 1} \right)\left( {3 - 1} \right)....\left( {n - 1} \right) + \left( {1 - 1} \right)\left( {3 - 1} \right)....\left( {n - 1} \right) + .... + \left( {1 - 1} \right)\left( {2 - 1} \right)....\left( {\left( {n - 1} \right) - 1} \right)} \right]$
Simplifying the calculations, we get,
$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left[ {\left( {2 - 1} \right)\left( {3 - 1} \right)....\left( {n - 1} \right) + \left( 0 \right)\left( {3 - 1} \right)....\left( {n - 1} \right) + .... + \left( 0 \right)\left( {2 - 1} \right)....\left( {\left( {n - 1} \right) - 1} \right)} \right]$
$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left( 1 \right)\left( 2 \right)....\left( {n - 1} \right)$
Condensing into factorial form, we get,
$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = \left( { - 1} \right)\left( {n - 1} \right)!$
So, the derivative of $y = \left( {1 - x} \right)\left( {2 - x} \right)....\left( {n - x} \right)$ with respect to $x$ is $\left( { - 1} \right)\left( {n - 1} \right)!$.

Hence, option B is the correct answer.

Note:
The derivatives of basic functions such as logarithm must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. Factorial is the expression for denoting the multiplication of consecutive integers.