Question

If $y=(x+1)(x+2)(x+3)(x+4)(x+5)$, then the value of $\frac{dy}{dx}$ at $x=0$is equal to

• A. $742$
• B. $472$
• C. $374$
• D. $274$

Answer 1

The correct option is D

$274$

Explanation for the correct option.

Given that, $y=(x+1)(x+2)(x+3)(x+4)(x+5)$

Differentiate both sides using product rule i.e;

$\frac{d\left(uv\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$

we have,$\frac{dy}{dx}=(\mathrm{x}+2)(\mathrm{x}+3)(\mathrm{x}+4)(\mathrm{x}+5)+(\mathrm{x}+1)(\mathrm{x}+3)(\mathrm{x}+4)(\mathrm{x}+5)+(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+4)(\mathrm{x}+5)+(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+3)(\mathrm{x}+5)+(\mathrm{x}+1)(\mathrm{x}+2)(x+3)(\mathrm{x}+4)$Now substitute $x=0$ in $\frac{dy}{dx}$we have:

$\begin{array}{rcl}\frac{dy}{dx}& =& (2\times 3\times 4\times 5)+(1\times 3\times 4\times 5)+(1\times 2\times 4\times 5)+(1\times 2\times 3\times 5)+(1\times 2\times 3\times 4)\\ & =& 120+60+40+30+24\\ & =& 274\end{array}$
Hence, option D is correct.