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If $y=\sin \left(m \sin ^{-1} x\right)$, what is the value of $d^{2} y / d x^{2}$ at $x=0$?
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$y=\sin \left(m \sin ^{-1} x\right)$
Then, $\frac{d y}{d x}=\cos \left(m \sin ^{-1} x\right) \frac{m}{\sqrt{1-x^{2}}}$
$\therefore \frac{d^{2} y}{d x^{2}}=\cos \left(m \sin ^{-1} x\right) \cdot m\left\{\frac{-1}{2} \cdot \frac{(-2 x)}{\left(1-x^{2}\right)^{3 / 2}}\right\}$
$+\frac{m}{\sqrt{1-x^{2}}}\left\{-\sin \left(m \sin ^{-1} x\right)\right\} \cdot \frac{m}{\sqrt{1-x^{2}}}$
$=\frac{m}{\sqrt{1-x^{2}}}\left[\frac{x}{\left(1-x^{2}\right)} \cos \left(m \sin ^{-1} x\right)\right]$
$\left.-\frac{m}{\sqrt{1-x^{2}}} \sin \left(m \sin ^{-1} x\right)\right]$
Now, $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is $m[0-0]=0 \quad\left(\because \sin ^{-1} 0=0\right)$
Then, $\frac{d y}{d x}=\cos \left(m \sin ^{-1} x\right) \frac{m}{\sqrt{1-x^{2}}}$
$\therefore \frac{d^{2} y}{d x^{2}}=\cos \left(m \sin ^{-1} x\right) \cdot m\left\{\frac{-1}{2} \cdot \frac{(-2 x)}{\left(1-x^{2}\right)^{3 / 2}}\right\}$
$+\frac{m}{\sqrt{1-x^{2}}}\left\{-\sin \left(m \sin ^{-1} x\right)\right\} \cdot \frac{m}{\sqrt{1-x^{2}}}$
$=\frac{m}{\sqrt{1-x^{2}}}\left[\frac{x}{\left(1-x^{2}\right)} \cos \left(m \sin ^{-1} x\right)\right]$
$\left.-\frac{m}{\sqrt{1-x^{2}}} \sin \left(m \sin ^{-1} x\right)\right]$
Now, $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is $m[0-0]=0 \quad\left(\because \sin ^{-1} 0=0\right)$
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