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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY2 Marks
Question
Differentiate w.r.t. x the function in Exercise:
$\sin^{-1}(\text{x}\sqrt{\text{x)}},\ 0\leq\text{x}\leq1$
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Answer

Let $\text{y}=\sin^{-1}(\text{x}\sqrt{\text{x)}}$
Using chain rule, we obtain
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin^{-1}(\text{x}\sqrt{\text{x})}$
$=\frac{1}{\sqrt{1-(\text{x}\sqrt{\text{x})^2}}}\times\frac{\text{d}}{\text{dx}}(\text{x}\sqrt{\text{x}})$
$=\frac{1}{\sqrt{1-(\text{x}\sqrt{\text{x})^2}}}.\frac{\text{d}}{\text{dx}}\Big(\text{x}^{\frac{3}{2}}\Big)$
$=\frac{1}{\sqrt{1-(\text{x}\sqrt{\text{x})^2}}}\times\frac{3}{2}.\text{x}^{\frac{1}{2}}$
$=\frac{3\sqrt{\text{x}}}{2\sqrt{1-\text{x}^3}}$
$=\frac{3}{2}\sqrt{\frac{\text{x}}{1-\text{x}^3}}$

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