Question

If $y=\frac{(x-2{)}^2(x+1{)}^3(x-3{)}^2}{(x+2)(x+4)},$ then the value of $\left(\frac{15}{32}\frac{dy}{dx}\right)$ at $x=1$ is

• A. $\frac{61}{30}$
• B. $-\frac{67}{30}$
• C. $-\frac{61}{30}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is C $-\frac{61}{30}$

Given :
$y=\frac{(x-2{)}^2(x+1{)}^3(x-3{)}^2}{(x+2)(x+4)}$
At $x=1$
$\Rightarrow y=\frac{1\times 8\times 4}{3\times 5}=\frac{32}{15}$
Taking $\ln$ on both sides,
$\ln y=2\ln (x-2)+3\ln (x+1)+2\ln (x-3)-\ln (x+2)-\ln (x+4)$
Differentiating w.r.t. $x,$ we get
$\frac{1}{y}\frac{dy}{dx}=\frac{2}{x-2}+\frac{3}{x+1}+\frac{2}{x-3}-\frac{1}{x+2}-\frac{1}{x+4}$
At $x=1$
$\frac{15}{32}\frac{dy}{dx}=[-\frac{2}{1}+\frac{3}{2}-\frac{2}{2}-\frac{1}{3}-\frac{1}{5}]$
$\Rightarrow \frac{15}{32}\frac{dy}{dx}=-\frac{61}{30}$