If y = frac{2(x - sin x)^{3/2}}{sqrt{x}},then | vedclass

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(B) Given $y = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}}$.Taking the natural logarithm on both sides:$\ln y = \ln 2 + \frac{3}{2} \ln(x - \sin x) - \frac{1

Vedclass · Accepted Answer

$\frac{2(x - \sin x)^{3/2}}{\sqrt{x}}\left[ \frac{3}{2} \cdot \frac{1 - \cos x}{x - \sin x} - \frac{1}{2x} \right]$

Vedclass · Answer

(B) Given $y = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}}$.Taking the natural logarithm on both sides:$\ln y = \ln 2 + \frac{3}{2} \ln(x - \sin x) - \frac{1}{2} \ln x$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = 0 + \frac{3}{2} \cdot \frac{1}{x - \sin x} \cdot (1 - \cos x) - \frac{1}{2} \cdot \frac{1}{x}$.Multiplying by $y$:$\frac{dy}{dx} = y \left[ \frac{3}{2} \cdot \frac{1 - \cos x}{x - \sin x} - \frac{1}{2x} \right]$.Substituting $y$ back:$\frac{dy}{dx} = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}} \left[ \frac{3}{2} \cdot \frac{1 - \cos x}{x - \sin x} - \frac{1}{2x} \right]$.

If $y = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}}$,then $\frac{dy}{dx} = $

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