Question:
$\sin ^{-1}(x \sqrt{x}), 0 \leq x \leq 1$
Solution:
Let $y=\sin ^{-1}(x \sqrt{x})$
Using chain rule, we obtain
$\frac{d y}{d x}=\frac{d}{d x} \sin ^{-1}(x \sqrt{x})$
$=\frac{1}{\sqrt{1-(x \sqrt{x})^{2}}} \times \frac{d}{d x}(x \sqrt{x})$
$=\frac{1}{\sqrt{1-x^{3}}} \cdot \frac{d}{d x}\left(x^{\frac{3}{2}}\right)$
$=\frac{1}{\sqrt{1-x^{3}}} \times \frac{3}{2} \cdot x^{\frac{1}{2}}$
$=\frac{3 \sqrt{x}}{2 \sqrt{1-x^{3}}}$
$=\frac{3}{2} \sqrt{\frac{x}{1-x^{3}}}$
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