Derivative of sin ^{-1}left(frac{t}{sqrt{1+t^ | vedclass

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(A) Let $y = \sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}ight)$.Put $t = 	an 	heta$,then $	heta = 	an ^{-1} t$.$y = \sin ^{-1}\left(\frac{	an 	heta}

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(A) Let $y = \sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}ight)$.Put $t = 	an 	heta$,then $	heta = 	an ^{-1} t$.$y = \sin ^{-1}\left(\frac{	an 	heta}{\sqrt{1+	an ^2 	heta}}ight) = \sin ^{-1}\left(\frac{	an 	heta}{\sec 	heta}ight) = \sin ^{-1}(\sin 	heta) = 	heta = 	an ^{-1} t$.Let $z = \cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}ight)$.$z = \cos ^{-1}\left(\frac{1}{\sqrt{1+	an ^2 	heta}}ight) = \cos ^{-1}\left(\frac{1}{\sec 	heta}ight) = \cos ^{-1}(\cos 	heta) = 	heta = 	an ^{-1} t$.Since $y = 	heta$ and $z = 	heta$,we have $y = z$.Therefore,$\frac{dy}{dz} = \frac{d}{dz}(z) = 1$.

Derivative of $\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$ with respect to $\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ is

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