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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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