Question

The value of $\frac{d}{dx}\left\{x\right(x-1\left)\right(x-2\left)\right(x-3\left)\right\}$ at $x=3$

• A. $0$
• B. $3$
• C. $6$
• D. $24$

Answer 1

Answer: (C)

$6$

Now,

$\frac{d}{dx}\left\{x\right(x-1\left)\right(x-2\left)\right(x-3\left)\right\}$

$=(x-3)\frac{d}{dx}\left\{x\right(x-1\left)\right(x-2\left)\right\}+$$x(x-1)(x-2)\frac{d}{dx}\left\{\right(x-3\left)\right\}$ [ Using product rule of differentiation ]

$=(x-3)\frac{d}{dx}\left\{x\right(x-1\left)\right(x-2\left)\right\}+$$x(x-1)(x-2).1$

At $x=3$ the value of $\frac{d}{dx}\left\{x\right(x-1\left)\right(x-2\left)\right(x-3\left)\right\}$ is $0$

$=0+\mathrm{3.2.1.1}$

$=6$.