Question

If $-1\le x\le 1$ then $\frac{d}{dx}\left({\sin }^{-1}(\frac{3x}{2}-\frac{x^3}{2})\right)$ is equal to

• A. $\frac{-3}{\sqrt{4-x^2}}$
• B. $\frac{3}{\sqrt{4-x^2}}$
• C. $\frac{1}{\sqrt{4-x^2}}$
• D. $\frac{-1}{\sqrt{4-x^2}}$

Answer 1

The correct option is A $\frac{-3}{\sqrt{4-x^2}}$

To find: $\frac{d}{dx}\left({\sin }^{-1}(\frac{3x}{2}-\frac{x^3}{2})\right)$

${\sin }^{-1}(\frac{3x}{2}-\frac{x^3}{2})={\sin }^{-1}\left(\frac{3x-x^3}{2}\right)$

$\frac{d}{dx}\left({\sin }^{-1}\left(\frac{3x-x^3}{2}\right)\right)$$=\frac{d}{du}\left({\sin }^{-1}\left(u\right)\right)\frac{d}{dx}\left(\frac{3x-x^3}{2}\right)$

$\frac{d}{du}\left({\sin }^{-1}\left(u\right)\right)$$=\frac{1}{\sqrt{1-u^2}}$

$\frac{d}{dx}\left(\frac{3x-x^3}{2}\right)$$=\frac{1}{2}\frac{d}{dx}(3x-x^3)$

$=\frac{1}{2}(\frac{d}{dx}\left(3x\right)-\frac{d}{dx}\left(x^3\right))$

$=\frac{3-3x^2}{2}$

$\therefore \frac{d}{dx}\left({\sin }^{-1}(\frac{3x}{2}-\frac{x^3}{2})\right)$$=\frac{1}{\sqrt{1-u^2}}\cdot \frac{3-3x^2}{2}$

$=\frac{1}{\sqrt{1-{\left(\frac{3x-x^3}{2}\right)}^2}}\cdot \frac{3-3x^2}{2}$

$=\frac{1}{\frac{\sqrt{{(x+1)}^2}\sqrt{{(x-1)}^2}\sqrt{-(x+2)(x-2)}}{2}}\cdot \frac{3-3x^2}{2}$

$=\frac{2}{\sqrt{{(x+1)}^2}\sqrt{{(x-1)}^2}\sqrt{-(x+2)(x-2)}}\cdot \frac{-3x^2+3}{2}$

$=\frac{3-3x^2}{\sqrt{{(x+1)}^2}\sqrt{{(x-1)}^2}\sqrt{-(x+2)(x-2)}}$

$=\frac{3(1-x^2)}{-1(1-x^2)\left(\sqrt{4-x^2}\right)}=\frac{-3}{\sqrt{4-x^2}}$