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If $\mathrm{u}=\sin ^{-1}(\mathrm{x}-\mathrm{y}), \mathrm{x}=3 \mathrm{t}, \mathrm{y}=4 \mathrm{t}^{3}$, then what is the derivative
of $\mathrm{u}$ with respect to $\mathrm{t}$?
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of $\mathrm{u}$ with respect to $\mathrm{t}$?
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Verified Answer
The correct answer is:
$3\left(1-t^{2}\right)^{-\frac{1}{2}}$
$\mathrm{u}=\sin ^{-1}(\mathrm{x}-\mathrm{y})$ and $\mathrm{x}=3 \mathrm{t}, \mathrm{y}=4 \mathrm{t}^{3}$
So, $\mathrm{u}=\sin ^{-1}\left(3 \mathrm{t}-4 \mathrm{t}^{3}\right)$
Let $\mathrm{t}=\sin \theta \Rightarrow \theta=\sin ^{-1} \mathrm{t}$
So, $\mathrm{u}=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right)$
$\quad=\sin ^{-1}(\sin 3 \theta)=3 \theta$
Hence, $\mathrm{u}=3 \sin ^{-1} \mathrm{t}$
$\frac{\mathrm{du}}{\mathrm{dt}}=3 \frac{1}{\sqrt{1-\mathrm{t}^{2}}}=3\left(1-\mathrm{t}^{2}\right)^{-1 / 2}$
So, $\mathrm{u}=\sin ^{-1}\left(3 \mathrm{t}-4 \mathrm{t}^{3}\right)$
Let $\mathrm{t}=\sin \theta \Rightarrow \theta=\sin ^{-1} \mathrm{t}$
So, $\mathrm{u}=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right)$
$\quad=\sin ^{-1}(\sin 3 \theta)=3 \theta$
Hence, $\mathrm{u}=3 \sin ^{-1} \mathrm{t}$
$\frac{\mathrm{du}}{\mathrm{dt}}=3 \frac{1}{\sqrt{1-\mathrm{t}^{2}}}=3\left(1-\mathrm{t}^{2}\right)^{-1 / 2}$
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