Question

The derivative of ${\sin }^{-1}(3x-4x^3)$ with respect to ${\sin }^{-1}x$ is

• A. $3,\text{for}|x|<1$
• B. $3,\text{for}|x|<\frac{1}{2}\text{and}-3\text{for}\frac{1}{2}<|x|<1$
• C. $-3,\text{for}|x|<1$
• D. $-3,\text{for}|x|\le \frac{1}{2}\text{and}3\text{for}\frac{1}{2}<|x|<1$

Answer 1

Answer: (B)

$3,\text{for}|x|<\frac{1}{2}\text{and}-3\text{for}\frac{1}{2}<|x|<1$

Let $y={\sin }^{-1}(3x-4x^3)$
$y$ can be defined in the interval as,
$y={\sin }^{-1}(3x-4x^3)=\{\begin{array}{ccc}3{\sin }^{-1}x& & ,\text{if}|x|\le \frac{1}{2}\\ \pi -3{\sin }^{-1}x& & ,\text{if}\frac{1}{2}<x\le 1\\ -\pi -3{\sin }^{-1}x& & ,\text{if}-1\le x<-\frac{1}{2}\end{array}$
Let $z={\sin }^{-1}x$. Then,
Putting this value of $z$ in the interval of $y$, we get
$y=\{\begin{array}{ccc}3z& & ,\text{if}|x|\le \frac{1}{2}\\ \pi -3z& & ,\text{if}\frac{1}{2}<x\le 1\\ -\pi -3z& & ,\text{if}-1\le x<-\frac{1}{2}\end{array}$
On differentiating, we get
$\Rightarrow \frac{dy}{dz}=\{\begin{array}{ccc}3& & ,\text{if}|x|<\frac{1}{2}\\ -3& & ,\text{if}\frac{1}{2}<|x|<1\end{array}$