The derivative of y = (1 – x) (2 – x)…. (n – x) at x = 1 is equal to

The correct option is (-1) (n – 1)!.

Let us write down the given function:
$$y=(1-x)(2-x)\cdots (n-x)$$

To find the derivative of this product of n linear terms, we can apply the product rule of differentiation. The product rule states that the derivative of a product of functions is the sum of the derivatives of each function multiplied by the remaining functions.
Let us write the function as:
$$y=u(x)\cdot v(x)$$
where
$u(x)=1-x$
and
$v(x)=(2-x)(3-x)\cdots (n-x)$

Differentiating both sides with respect to x using the product rule:
$$\frac{dy}{dx}=u^{\prime }(x)\cdot v(x)+u(x)\cdot v^{\prime }(x)$$

First, find the derivative of
$u(x)=1-x$:
$$u^{\prime }(x)=\frac{d}{dx}(1-x)=-1$$

Now substitute this back into our expression for the derivative:
$$\frac{dy}{dx}=(-1)\cdot [(2-x)(3-x)\cdots (n-x)]+(1-x)\cdot v^{\prime }(x)$$

We need to evaluate the value of the derivative at
$x=1$. Let us substitute
$x=1$ into the derivative expression:
$${\left[\frac{dy}{dx}\right]}_{x=1}=(-1)\cdot [(2-1)(3-1)\cdots (n-1)]+(1-1)\cdot v^{\prime }(1)$$

Notice that the second term containing
$(1-1)$ becomes zero:
$$(1-1)\cdot v^{\prime }(1)=0\cdot v^{\prime }(1)=0$$

Thus, we are left with only the first term:
$${\left[\frac{dy}{dx}\right]}_{x=1}=(-1)\cdot [(1)\cdot (2)\cdot (3)\cdots (n-1)]$$

By definition, the product of consecutive integers from 1 up to
$n-1$ is the factorial of
$n-1$, which is denoted by
$(n-1)!$.
Substituting this factorial notation in, we obtain:
$${\left[\frac{dy}{dx}\right]}_{x=1}=(-1)(n-1)!$$