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Differentiate w.r.t. \(x\) the function in \( sin^{-1}(x \: \sqrt x ), 0 \leq x \leq 1 \)

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Toolbox:
  • $\large\frac{d}{dx}$$(\sin^{-1}x)=\large\frac{1}{\sqrt{1-x^2}}$
  • $\large\frac{dy}{dx}=\large\frac{dy}{dt}$$\times \large\frac{dt}{dx}$
Step 1:
Let $y=\sin^{-1}(x\sqrt{x})$
$\qquad=\sin^{-1}(x.x^{\large\frac{1}{2}})$
$\qquad=\sin^{-1}(x^{\large\frac{3}{2}})$
Put $x^{\Large\frac{3}{2}}=t$
$y=\sin^{-1}t$
Differentiating with respect to $t$
$\large\frac{dy}{dt}=\frac{1}{\sqrt{1-t^2}}$
$\large\frac{dt}{dx}$$=\large\frac{3}{2}$$x^{\Large\frac{3}{2}-\normalsize1}$
$\quad=\large\frac{3}{2}$$x^{\Large\frac{1}{2}}$
Step 2:
$\large\frac{dy}{dx}=\large\frac{dy}{dt}$$\times \large\frac{dt}{dx}$
$\quad\;=\large\frac{1}{\sqrt{1-t^2}}$$\times \large\frac{3}{2}$$x^{\Large\frac{1}{2}}$
$\quad\;=\large\frac{1}{\sqrt{1-t^2}}$$\times \large\frac{3}{2}$$\sqrt x$
$\quad\;=\large\frac{3.\sqrt x}{2\sqrt{1-t^2}}$
answered May 14, 2013 by sreemathi.v
 

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