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If $u=\sin \left(\frac{x}{y}\right), x=e^t$ and $y=t^2$, then$t^6\left(\frac{d u}{d t}\right)^2 \div e^{2 t}(t-2)^2=$
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Verified Answer
The correct answer is:
$1-u^2$
Given $u=\sin \left(\frac{x}{y}\right), x=e^t, y=t^2 \Rightarrow u=\sin \left(\frac{e^t}{t^2}\right)$
Now, $\frac{d u}{d t}=\cos \left(\frac{e^t}{t^2}\right)\left\{\frac{t^2 e^t-e^t \cdot 2 t}{\left(t^2\right)^2}\right\}$
$\begin{aligned}
& =\cos \left(\frac{e^t}{t^2}\right)\left(\frac{t-2}{t^3}\right) e^t \\
& \Rightarrow\left(\frac{d u}{d t}\right)^2=\cos ^2\left(\frac{e^t}{t^2}\right) \frac{(t-2)^2}{t^6} e^{2 t} \\
& \Rightarrow t^6\left(\frac{d u}{d t}\right)^2 \div(t-2)^2 e^{2 t} \\
& =\cos ^2\left(\frac{e^t}{t^2}\right)=1-\sin ^2\left(\frac{e^t}{t^2}\right)=1-u^2
\end{aligned}$
Now, $\frac{d u}{d t}=\cos \left(\frac{e^t}{t^2}\right)\left\{\frac{t^2 e^t-e^t \cdot 2 t}{\left(t^2\right)^2}\right\}$
$\begin{aligned}
& =\cos \left(\frac{e^t}{t^2}\right)\left(\frac{t-2}{t^3}\right) e^t \\
& \Rightarrow\left(\frac{d u}{d t}\right)^2=\cos ^2\left(\frac{e^t}{t^2}\right) \frac{(t-2)^2}{t^6} e^{2 t} \\
& \Rightarrow t^6\left(\frac{d u}{d t}\right)^2 \div(t-2)^2 e^{2 t} \\
& =\cos ^2\left(\frac{e^t}{t^2}\right)=1-\sin ^2\left(\frac{e^t}{t^2}\right)=1-u^2
\end{aligned}$
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