Question

Let $y={\sin }^{-1}x$, then find $(1-x^2)y_2-x{y}_1$.
Where $y_1$ and $y_2$ denote the first and second order derivatives respectively.

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (C)

$0$

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

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

$y_2=\frac{d^{2}y}{d{x}^2}=-\frac{1}{2}\frac{1}{{(1-x^2)}^{\frac{3}{2}}}(-2x)=\frac{x}{(1-x^2)\sqrt{1-x^2}}$

Thus the value of $(1-x^2)y_2-x{y}_1$ is

$(1-x^2)\frac{x}{(1-x^2)\sqrt{1-x^2}}-x\frac{1}{\sqrt{1-x^2}}$

$=\frac{x}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}=0$

So, option C is correct