If f'(x) = sin^2 x and y = f left( frac{2x - | vedclass

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(D) Let $u = \frac{2x - 1}{x^2 + 1}$. Then $y = f(u)$, so $\frac{dy}{dx} = f'(u) \cdot \frac{du}{dx}$.First, calculate $\frac{du}{dx}$ using the quoti

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

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(D) Let $u = \frac{2x - 1}{x^2 + 1}$. Then $y = f(u)$, so $\frac{dy}{dx} = f'(u) \cdot \frac{du}{dx}$.First, calculate $\frac{du}{dx}$ using the quotient rule: $\frac{du}{dx} = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2}$.At $x = 1$, $u = \frac{2(1) - 1}{1^2 + 1} = \frac{1}{2}$ and $\frac{du}{dx} = \frac{-2(1)^2 + 2(1) + 2}{(1^2 + 1)^2} = \frac{2}{4} = \frac{1}{2}$.Since $f'(x) = \sin^2 x$, we have $f'(u) = \sin^2 u$. Thus, $f'(1/2) = \sin^2(1/2)$.Therefore, $\frac{dy}{dx} = f'(1/2) \cdot \frac{du}{dx} = \sin^2(1/2) \cdot \frac{1}{2} = \frac{1}{2} \sin^2 \left( \frac{1}{2} \right)$.

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

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