Question

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

• A. $\frac{11}{8}$
• B. $\frac{13}{8}$
• C. $\frac{31}{8}$
• D. $\frac{17}{8}$

Answer 1

The correct option is C $\frac{31}{8}$

Given :
$y=\frac{(x-1)(x+2)(x-4)}{(x-2)(x+1)}$
At $x=3$
$\Rightarrow y=\frac{2\times 5\times -1}{1\times 4}=-\frac{5}{2}$
Taking $\ln$ on both sides,
$\ln y=\ln (x-1)+\ln (x+2)+\ln (x-4)-\ln (x-2)-\ln (x+1)$
Differentiating w.r.t. $x,$ we get
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x-1}+\frac{1}{x+2}+\frac{1}{x-4}-\frac{1}{x-2}-\frac{1}{x+1}$
At $x=3$
$\frac{dy}{dx}=-\frac{5}{2}[\frac{1}{2}+\frac{1}{5}-\frac{1}{1}-\frac{1}{1}-\frac{1}{4}]$
$\Rightarrow \frac{dy}{dx}=-\frac{5}{2}\times -\frac{31}{20}=\frac{31}{8}$