If y = f left( frac{2x - 1}{x^2 + 1} right) a | vedclass

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(B) Given $y = f \left( \frac{2x - 1}{x^2 + 1} \right)$ and $f'(x) = \sin x$. Applying the chain rule,we differentiate $y$ with respect to $x$: $\frac

Vedclass · Accepted Answer

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

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(B) Given $y = f \left( \frac{2x - 1}{x^2 + 1} \right)$ and $f'(x) = \sin x$. Applying the chain rule,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = f' \left( \frac{2x - 1}{x^2 + 1} \right) \cdot \frac{d}{dx} \left( \frac{2x - 1}{x^2 + 1} \right)$. Since $f'(x) = \sin x$,we have $f' \left( \frac{2x - 1}{x^2 + 1} \right) = \sin \left( \frac{2x - 1}{x^2 + 1} \right)$. Now,differentiate $u = \frac{2x - 1}{x^2 + 1}$ using the quotient rule: $\frac{du}{dx} = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{2 + 2x - 2x^2}{(x^2 + 1)^2} = \frac{2(1 + x - x^2)}{(x^2 + 1)^2}$. Substituting these back into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2(1 + x - x^2)}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)$. Thus,the correct option is $B$.

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

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