Question

If $y={\sin }^{-1}\left(3x\right)+{\sec }^{-1}\left(\frac{1}{3x}\right)$, find $\frac{dy}{dx}$

Answer 1

Given equation

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

$⟹y={\sin }^{-1}\left(3x\right)+{\cos }^{-1}\left(3x\right)$

${\sin }^{-1}x+{\cos }^{1}x=\frac{\pi }{2}$

$⟹y=\frac{\pi }{2}$

$⟹\frac{dy}{dx}=0$