Question

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

[1 mark]

• A. $0$
• B. $2$
• C. $6$
• D. $5$

Answer 1

The correct option is C $6$

$y=(1+x)(1+2x)(1+3x)$
At $x=0,y=1$
Taking $\ln$ on both sides, we get
$\ln y=\ln (1+x)+\ln (1+2x)+\ln (1+3x)$
Differentiating w.r.t. $x,$ we get
$\frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{1+x}+\frac{2}{1+2x}+\frac{3}{1+3x}$
$\Rightarrow \frac{dy}{dx}=y(\frac{1}{1+x}+\frac{2}{1+2x}+\frac{3}{1+3x})$
$\Rightarrow \frac{dy}{dx}{|}_{(0,1)}=1(1+2+3)=6$