If y = sin(m sin^{-1} x),then (1 - x^2) y_2 - | vedclass

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(B) Given $y = \sin(m \sin^{-1} x)$.Differentiating with respect to $x$:$y_1 = \cos(m \sin^{-1} x) \cdot \frac{m}{\sqrt{1 - x^2}}$$y_1 \sqrt{1 - x^2}

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$-m^2 y$

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(B) Given $y = \sin(m \sin^{-1} x)$.Differentiating with respect to $x$:$y_1 = \cos(m \sin^{-1} x) \cdot \frac{m}{\sqrt{1 - x^2}}$$y_1 \sqrt{1 - x^2} = m \cos(m \sin^{-1} x)$Squaring both sides:$y_1^2 (1 - x^2) = m^2 \cos^2(m \sin^{-1} x) = m^2 (1 - \sin^2(m \sin^{-1} x)) = m^2 (1 - y^2)$$y_1^2 (1 - x^2) = m^2 - m^2 y^2$Differentiating again with respect to $x$:$2 y_1 y_2 (1 - x^2) + y_1^2 (-2x) = -2 m^2 y y_1$Dividing by $2 y_1$ (assuming $y_1 
eq 0$):$y_2 (1 - x^2) - x y_1 = -m^2 y$

If $y = \sin(m \sin^{-1} x)$,then $(1 - x^2) y_2 - x y_1$ is equal to (Here,$y_n$ denotes $\frac{d^n y}{dx^n}$)

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