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

Given the function $y = \frac{K^{\cos^{-1} x}}{1 + K^{\cos^{-1} x}}$ and the substitution $t = K^{\cos^{-1} x}$, we find $\frac{dy}{dt}$.

Substituting $t$ into the expression for $y$ yields $y = \frac{t}{1 + t}$.

Differentiating with respect to $t$ using the quotient rule gives:

$\frac{dy}{dt} = \frac{(1)(1 + t) - t(1)}{(1 + t)^2} = \frac{1}{(1 + t)^2}$.

Replacing $t$ with $K^{\cos^{-1} x}$ provides the final derivative:

$\frac{dy}{dt} = \frac{1}{(1 + K^{\cos^{-1} x})^2}$,

which corresponds to option C.