Question

If $y={\left(\frac{x^2}{x+1}\right)}^x$ then $\frac{dy}{dx}$ is

• A. $y[\frac{x+2}{x+1}+\log \left(\frac{x+1}{x^2}\right)]$
• B. $y[\frac{x+1}{x+2}+\log \left(\frac{x^2}{x+1}\right)]$
• C. $y[\frac{x+2}{x+1}+\log \left(\frac{x^2}{x+1}\right)]$
• D. $Noneofthese$

Answer 1

Answer: (C)

$y[\frac{x+2}{x+1}+\log \left(\frac{x^2}{x+1}\right)]$

Consider the following differential eq.

$y={\left(\frac{x^2}{x+1}\right)}^x$

Taking $log$ both side, we get .

$\log y=x\log \left(\frac{x^2}{x+1}\right)$

On differentiating both side w.r.t $x$

$\frac{1}{y}\frac{dy}{dx}=1\times \log \left(\frac{x^2}{x+1}\right)+x\left(\frac{(x+1)}{x^2}\right)\left(\frac{2x(x+1)-x^2}{{(x+1)}^2}\right)$

$\frac{1}{y}\frac{dy}{dx}=\log \left(\frac{x^2}{x+1}\right)+\frac{x+2}{x+1}$

$\frac{dy}{dx}=y(\log \left(\frac{x^2}{x+1}\right)+\frac{x+2}{x+1})$

Hence, this is the required answer.