If y = frac{a^{cos^{-1}x}}{1 + a^{cos^{-1}x}} | vedclass

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(C) Given $y = \frac{a^{\cos^{-1}x}}{1 + a^{\cos^{-1}x}}$ and $z = a^{\cos^{-1}x}$.Substituting $z$ into the expression for $y$,we get $y = \frac{z}{1

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$\frac{1}{(1 + a^{\cos^{-1}x})^2}$

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(C) Given $y = \frac{a^{\cos^{-1}x}}{1 + a^{\cos^{-1}x}}$ and $z = a^{\cos^{-1}x}$.Substituting $z$ into the expression for $y$,we get $y = \frac{z}{1 + z}$.Now,differentiate $y$ with respect to $z$ using the quotient rule $\frac{d}{dz}(\frac{u}{v}) = \frac{v \frac{du}{dz} - u \frac{dv}{dz}}{v^2}$.$\frac{dy}{dz} = \frac{(1 + z)(1) - z(1)}{(1 + z)^2}$.$\frac{dy}{dz} = \frac{1 + z - z}{(1 + z)^2} = \frac{1}{(1 + z)^2}$.Substituting back $z = a^{\cos^{-1}x}$,we get $\frac{dy}{dz} = \frac{1}{(1 + a^{\cos^{-1}x})^2}$.

If $y = \frac{a^{\cos^{-1}x}}{1 + a^{\cos^{-1}x}}$ and $z = a^{\cos^{-1}x}$,then $\frac{dy}{dz} = $

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