A function y = f(x) is given by x = frac{1}{1 | vedclass

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(B) Given $x = (1 + t^2)^{-1}$ and $y = \frac{1}{t(1 + t^2)}$.First,we find $\frac{dx}{dt} = -1(1 + t^2)^{-2} \cdot (2t) = -\frac{2t}{(1 + t^2)^2}$.Ne

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increasing in $(0, 1)$

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(B) Given $x = (1 + t^2)^{-1}$ and $y = \frac{1}{t(1 + t^2)}$.First,we find $\frac{dx}{dt} = -1(1 + t^2)^{-2} \cdot (2t) = -\frac{2t}{(1 + t^2)^2}$.Next,we find $\frac{dy}{dt}$ using the quotient rule or product rule: $y = t^{-1}(1 + t^2)^{-1}$.$\frac{dy}{dt} = -t^{-2}(1 + t^2)^{-1} + t^{-1}(-1)(1 + t^2)^{-2}(2t) = -\frac{1}{t^2(1 + t^2)} - \frac{2}{(1 + t^2)^2} = -\frac{1 + t^2 + 2t^2}{t^2(1 + t^2)^2} = -\frac{1 + 3t^2}{t^2(1 + t^2)^2}$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-(1 + 3t^2) / [t^2(1 + t^2)^2]}{-2t / (1 + t^2)^2} = \frac{1 + 3t^2}{2t^3}$.Since $t > 0$,$1 + 3t^2 > 0$ and $2t^3 > 0$,so $\frac{dy}{dx} > 0$ for all $t > 0$.As $t 	o \infty$,$x 	o 0$,and as $t 	o 0$,$x 	o 1$. Thus,for $t \in (0, \infty)$,$x \in (0, 1)$.Since $\frac{dy}{dx} > 0$,the function $f$ is increasing in $(0, 1)$.

$A$ function $y = f(x)$ is given by $x = \frac{1}{1 + t^2}$ and $y = \frac{1}{t(1 + t^2)}$ for all $t > 0$. Then $f$ is:

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