Question

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

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

Answer 1

Answer: (B)

$\frac{2(1+x-x^2)}{(1+x^2{)}^2}\sin \left(\frac{2x-1}{x^2+1}\right)$

It is given that

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

using chain rule,

$\frac{dy}{dx}=f^{\prime }\left(\frac{2x-1}{x^2+1}\right)\frac{d\left(\frac{2x-1}{x^2+1}\right)}{dx}$

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

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