If y = fleft( frac{2x - 1}{x^2 + 1} right) an | vedclass

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(D) Given $y = f\left( \frac{2x - 1}{x^2 + 1} \right).$ Let $t = \frac{2x - 1}{x^2 + 1}.$By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}.

Vedclass · Accepted Answer

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

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(D) Given $y = f\left( \frac{2x - 1}{x^2 + 1} \right).$ Let $t = \frac{2x - 1}{x^2 + 1}.$By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}.$Since $f'(x) = \sin(x^2),$ we have $f'(t) = \sin(t^2) = \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$Now,calculate $\frac{dt}{dx}$ using the quotient rule:$\frac{dt}{dx} = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2}.$Therefore,$\frac{dy}{dx} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$

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

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