Question

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

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

Answer 1

Answer: (C)

$ \frac{dy}{dx} = f'\left(z\right) \frac{dz}{dx} \,where \, z = \frac{2x-1}{x^{2} + 1}$
$ = \sin z^{2} . \frac{\left(x^{2} + 1\right)2 - \left(2x -1\right) 2x}{\left(x^{2} + 1\right)^{2}} $
$=\sin\left(\frac{2x-1}{x^{2}+ 1}\right)^{2} . \frac{2+2x-2x^{2}}{\left(x^{2} + 1 \right)^{2}} $