Question

y=sin inverse x prove that d^2y/dx^2-x dy/dx = 0

Answer 1

$\textbf{Answer}$

$Given \: \: that, \\ \\ y = {sin}^{ - 1} x \\ \\ Differentiating \: \: both \: \: sides \\ with \: \: respect \: \: to \: \: x, \: \: we \: \: get - \\ \\ \frac{dy}{dx} = \frac{1}{ \sqrt{1 - {x}^{2} } } \\ \\ Or, \: \: \sqrt{1 - {x}^{2} } \: \: \frac{dy}{dx} = 1 \\ \\ Squaring \: \: both \: \: sides, \: \: we \: \: get- \\ \\ (1 - {x}^{2} ) {( \frac{dy}{dx} )}^{2} = 1 \\ \\ Again, \: \: differentiating \: \: with \\ respect \: \: to \: \: x, \: \: we \: \: get- \\ \\ 2(1 - {x}^{2} ) \frac{dy}{dx} \frac{ {d}^{2} y}{d {x}^{2} } - 2x { (\frac{dy}{dx}) }^{2} = 0 \\ \\ Or, \: \: (1 - {x}^{2} ) \frac{ {d}^{2}y }{d {x}^{2} } - x \frac{dy}{dx} = 0 \\ \\ Hence, \: \: proved.$

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