Question

If $y=f\left(\frac{2x-1}{x^2+1}\right)$ and $f^{\prime }\left(x\right)=\sin x^2$ then $\frac{dy}{dx}$ is equal to-

• A. $sin{\left(\frac{2x-1}{x^2+1}\right)}^2\left(\frac{2+2x+x^2}{{(x^2+1)}^2}\right)$
• B. $sin{\left(\frac{2x-1}{x^2+1}\right)}^2\left(\frac{2+2x-2x^2}{{(x^2+1)}^2}\right)$
• C. $sin{\left(\frac{2x-1}{x^2+1}\right)}^2\left(\frac{2+2x-x^2}{{(x^2+1)}^2}\right)$
• D. None of these

Answer 1

The correct option is B $sin{\left(\frac{2x-1}{x^2+1}\right)}^2\left(\frac{2+2x-2x^2}{{(x^2+1)}^2}\right)$

$y=f\left(\frac{2x-1}{x^2+1}\right)$ and $f^{\prime }\left(x\right)=\sin x^2$
$y^{\prime }=f^{\prime }\left(\frac{2x-1}{x^2+1}\right)\times \frac{2x^2+2-4x^2+2x}{(x^2+1{)}^2}$
$=f^{\prime }\left(\frac{2x-1}{x^2+1}\right)\frac{-2x^2+2x+2}{(x^2+1{)}^2}$
$=\sin {\left(\frac{2x-1}{x^2+1}\right)}^2\left(\frac{2+2x-2x^2}{(x^2+1{)}^2}\right)$ ($\because f^{\prime }\left(x\right)=\sin x^2$)