If u=sin left(frac{x}{y}right),x=e^t,and y=t^ | vedclass

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(C) Given $u=\sin \left(\frac{x}{y}\right)$,$x=e^t$,and $y=t^2$. Substituting $x$ and $y$ in $u$,we get $u=\sin \left(\frac{e^t}{t^2}\right)$.Now,diff

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$1-u^2$

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(C) Given $u=\sin \left(\frac{x}{y}ight)$,$x=e^t$,and $y=t^2$. Substituting $x$ and $y$ in $u$,we get $u=\sin \left(\frac{e^t}{t^2}ight)$.Now,differentiating $u$ with respect to $t$ using the chain rule:$\frac{d u}{d t}=\cos \left(\frac{e^t}{t^2}ight) \cdot \frac{d}{d t}\left(\frac{e^t}{t^2}ight)$Using the quotient rule for $\frac{d}{d t}\left(\frac{e^t}{t^2}ight) = \frac{t^2 e^t - e^t(2t)}{(t^2)^2} = \frac{t e^t(t-2)}{t^4} = \frac{e^t(t-2)}{t^3}$.Thus,$\frac{d u}{d t} = \cos \left(\frac{e^t}{t^2}ight) \cdot \frac{e^t(t-2)}{t^3}$.Squaring both sides:$\left(\frac{d u}{d t}ight)^2 = \cos^2 \left(\frac{e^t}{t^2}ight) \cdot \frac{e^{2t}(t-2)^2}{t^6}$.Rearranging the expression $t^6\left(\frac{d u}{d t}ight)^2 \div \left(e^{2 t}(t-2)^2ight)$:$= \frac{t^6 \cdot \cos^2 \left(\frac{e^t}{t^2}ight) \cdot e^{2t}(t-2)^2}{t^6 \cdot e^{2t}(t-2)^2} = \cos^2 \left(\frac{e^t}{t^2}ight)$.Since $u = \sin \left(\frac{e^t}{t^2}ight)$,we have $\sin^2 \left(\frac{e^t}{t^2}ight) = u^2$.Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get $\cos^2 \left(\frac{e^t}{t^2}ight) = 1 - u^2$.

If $u=\sin \left(\frac{x}{y}\right)$,$x=e^t$,and $y=t^2$,then $t^6\left(\frac{d u}{d t}\right)^2 \div \left(e^{2 t}(t-2)^2\right)=$

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