Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

Given $y=\frac{1}{2}(si{n}^{-1}x{)}^2$.

We have to find $(1-x^2)y_2-x{y}_1$ where $y_1,y_2$ denote the first and second derivatives $y$ respectively.

$y_1=\frac{d}{dx}\left(\frac{1}{2}\right(si{n}^{-1}x{)}^2)$

$=\frac{1}{2}\cdot \frac{d}{dx}{\left(si{n}^{-1}x\right)}^2$

$=\frac{1}{2}\cdot 2si{n}^{-1}x\cdot \frac{1}{\sqrt{1-x^2}}$

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

Now let us find $y_2$

$y_2=\frac{d}{dx}{y}_1$

$=\frac{d}{dx}\left(\frac{si{n}^{-1}x}{\sqrt{1-x^2}}\right)$

$=\frac{\sqrt{1-x^2}\cdot \frac{d}{dx}\left(si{n}^{-1}x\right)-si{n}^{-1}x\cdot \frac{d}{dx}\left(\sqrt{1-x^2}\right)}{(\sqrt{1-x^2}{)}^2}$

$=\frac{\sqrt{1-x^2}\cdot \frac{1}{\sqrt{1-x^2}}-si{n}^{-1}x\cdot (-\frac{x}{\sqrt{1-x^2}})}{(\sqrt{1-x^2}{)}^2}$

$=\frac{1+\frac{xsi{n}^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$

$\therefore y_2=\frac{\sqrt{1-x^2}+xsi{n}^{-1}x}{(1-x^2)\left(\sqrt{1-x^2}\right)}$

Now we have to find $(1-x^2)y_2-x{y}_1$.

$(1-x^2)y_2-x{y}_1=(1-x^2)\cdot \frac{\sqrt{1-x^2}+xsi{n}^{-1}x}{(1-x^2)\left(\sqrt{1-x^2}\right)}-x\cdot \frac{si{n}^{-1}x}{\sqrt{1-x^2}}$

$=\frac{\sqrt{1-x^2}+xsi{n}^{-1}x}{\sqrt{1-x^2}}-\frac{xsi{n}^{-1}x}{\sqrt{1-x^2}}$

$=1+\frac{xsi{n}^{-1}x}{\sqrt{1-x^2}}-\frac{xsi{n}^{-1}x}{\sqrt{1-x^2}}$

$\therefore (1-x^2)y_2-x{y}_1=1$