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Differentiation under Integral Sign

If y=√(x+1)-√...

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If y=√(x+1)-√(x-1) then prove that: (x²-1)d²y/dx²+x.dy/dx=¼y

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x \sqrt{1 + y} + y \sqrt{1 + x} = 0

x \neq y

\frac{d y}{d x} = - \frac{1}{(x + 1)^{2}}

x \sqrt{1 + y} + y \sqrt{1 + x} = 0, - 1 < x < 1, x \neq y

then prove that

\frac{d y}{d x} = - \frac{1}{{(1 + x)}^{2}}

y = \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .,

\frac{d y}{d x} =

y = {sin}^{- 1} x

, then prove that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 0

x = \frac{1 - t^{2}}{1 + t^{2}}, y = \frac{2 t}{1 + t^{2}}

, then prove that

\frac{d y}{d x} = - \frac{x}{y}

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