If x = frac{9t^2}{1+t^4} and y = frac{16t^2}{ | vedclass

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(D) Given $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$.First, differentiate $y$ with respect to $t$ using the quotient rule:$\frac{dy}{dt} =

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$\frac{16}{9}\left(\frac{1+t^4}{1-t^4}\right)^3$

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(D) Given $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$.First, differentiate $y$ with respect to $t$ using the quotient rule:$\frac{dy}{dt} = \frac{(1-t^4)(32t) - (16t^2)(-4t^3)}{(1-t^4)^2} = \frac{32t - 32t^5 + 64t^5}{(1-t^4)^2} = \frac{32t(1+t^4)}{(1-t^4)^2}$.Next, differentiate $x$ with respect to $t$ using the quotient rule:$\frac{dx}{dt} = \frac{(1+t^4)(18t) - (9t^2)(4t^3)}{(1+t^4)^2} = \frac{18t + 18t^5 - 36t^5}{(1+t^4)^2} = \frac{18t(1-t^4)}{(1+t^4)^2}$.Now, find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:$\frac{dy}{dx} = \frac{32t(1+t^4)}{(1-t^4)^2} 	imes \frac{(1+t^4)^2}{18t(1-t^4)} = \frac{32(1+t^4)^3}{18(1-t^4)^3} = \frac{16}{9}\left(\frac{1+t^4}{1-t^4}ight)^3$.

If $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$, then $\frac{dy}{dx} = $

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