Question

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

• A. $\left\{\frac{\left[6x^2-2x+2\right]}{{\left(x^2+1\right)}^2}\right\}\sin \left[\frac{(2x-1{)}^2}{\left(x^2+1\right)}\right]$
• B. $\left\{\frac{\left[6x^2-2x+2\right]}{{\left(x^2+1\right)}^2}\right\}{\sin }^2\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]$
• C. $\left\{\frac{\left[-2x^2+2x+2\right]}{{\left(x^2+1\right)}^2}\right\}{\sin }^2\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]$
• D. $\left\{\frac{\left[-2x^2+2x+2\right]}{{\left(x^2+1\right)}^2}\right\}\sin {\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]}^2$

Answer 1

The correct option is C

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

Explanation for the correct option.

Find the $\frac{dy}{dx}$.

Given that, $y=f\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]$ and $f^{\text{'}}\left(x\right)={\sin }^{2}x$.

Differentiate $y$ with respect to $x$using chain rule.

$\begin{array}{rcl}\frac{dy}{dx}& =& f^{\text{'}}\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]\frac{d\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]}{dx}\\ & =& {\sin }^2\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]\frac{[\left(x^2+1\right)2-(2x-1\left)\right(2x\left)\right]}{{\left(x^2+1\right)}^2}\\ & =& {\sin }^2\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]\left[\frac{\left(-2x^2+2x+2\right)}{{\left(x^2+1\right)}^2}\right]\\ & =& \left[\frac{-2x^2+2x+2}{{\left(x^2+1\right)}^2}\right]{\sin }^2\left[\frac{(2x-1)}{\left(x^2+1\right)}\right]\end{array}$

Hence, option C is correct.