Question

If $f\left(x\right)=(x+1)(x+2)(x+3).....(x+n)$, then find $f^{\prime }\left(0\right)$.

• A. $\left(n\right)!\{1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\}$
• B. $\left(n\right)!$
• C. $\left(n\right){!}^2\{1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\}$
• D. $n(n+1)/2$

Answer 1

The correct option is A $\left(n\right)!\{1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\}$

$f\left(x\right)=(x+1)(x+2)...(x+n)$

Taking log on both sides we get,

$logf\left(x\right)=log(x+1)+log(x+2)+...log(x+n)$

Differentiating with respect to $x$ we get,

$\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+n})$

At $x=0$, we have,

$\frac{f^{\prime }\left(0\right)}{f\left(0\right)}=(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n})$

$f\left(0\right)=n!$

$f^{\prime }\left(0\right)=n!(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n})$