If y = sin^{-1} left( frac{25 - x^2}{25 + x^2 | vedclass

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(A) Let $x = 5 	an 	heta$, then $	heta = 	an^{-1} \left( \frac{x}{5} ight)$.Substituting this into the expression: $y = \sin^{-1} \left( \frac{2

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$-\frac{5}{13}$

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(A) Let $x = 5 	an 	heta$, then $	heta = 	an^{-1} \left( \frac{x}{5} ight)$.Substituting this into the expression: $y = \sin^{-1} \left( \frac{25 - 25 	an^2 	heta}{25 + 25 	an^2 	heta} ight) = \sin^{-1} \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight) = \sin^{-1} (\cos 2	heta)$.Using the identity $\cos 2	heta = \sin \left( \frac{\pi}{2} - 2	heta ight)$, we get $y = \sin^{-1} \left( \sin \left( \frac{\pi}{2} - 2	heta ight) ight) = \frac{\pi}{2} - 2	heta$.Substituting back $	heta$: $y = \frac{\pi}{2} - 2 	an^{-1} \left( \frac{x}{5} ight)$.Differentiating with respect to $x$: $y' = 0 - 2 \cdot \frac{1}{1 + (x/5)^2} \cdot \frac{1}{5} = -\frac{2}{5} \cdot \frac{1}{1 + x^2/25} = -\frac{2}{5} \cdot \frac{25}{25 + x^2} = -\frac{10}{25 + x^2}$.At $x = 1$: $y'(1) = -\frac{10}{25 + 1^2} = -\frac{10}{26} = -\frac{5}{13}$.

If $y = \sin^{-1} \left( \frac{25 - x^2}{25 + x^2} \right)$, then find the value of $y'(1)$.

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