Question

Let $y={\sin }^{-1}x$ then prove that $(1-x^2)y_2-x{y}_1=0$. Where $y_1$ and $y_2$ are first and second order derivatives of $y$ with respect to $x$ respectively.

Answer 1

Given,

$y={\sin }^{-1}x$

Now differentiating both sides we get,

$y_1=\frac{1}{\sqrt{1-x^2}}$

Now squaring both sides we get,

$(1-x^2)y_1^2=1$

Now again differentiating both sides with respect to $x$ we get,

$(1-x^2)2y_1.y_2-2x.y_1^2=0$

or, $(1-x^2)y_2-x{y}_1=0$. [ Since $y_1\ne 0$]