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

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(C) Given that $t = K^{\cos^{-1} x}$.Substituting $t$ into the expression for $y$,we get $y = \frac{t}{1+t}$.To find $\frac{dy}{dt}$,we differentiate

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

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(C) Given that $t = K^{\cos^{-1} x}$.Substituting $t$ into the expression for $y$,we get $y = \frac{t}{1+t}$.To find $\frac{dy}{dt}$,we differentiate $y$ with respect to $t$ using the quotient rule:$\frac{dy}{dt} = \frac{d}{dt} \left( \frac{t}{1+t} \right)$.Using the quotient rule $\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$,where $u = t$ and $v = 1+t$:$\frac{dy}{dt} = \frac{(1+t)(1) - t(1)}{(1+t)^2}$.$\frac{dy}{dt} = \frac{1+t-t}{(1+t)^2} = \frac{1}{(1+t)^2}$.Substituting $t = K^{\cos^{-1} x}$ back into the expression,we get:$\frac{dy}{dt} = \frac{1}{(1 + K^{\cos^{-1} x})^2}$.

If $y = \frac{K^{\cos^{-1} x}}{1 + K^{\cos^{-1} x}}$ and $t = K^{\cos^{-1} x}$,then find $\frac{dy}{dt}$.

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