Question

If y = sec^-1 $\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+{\sin }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right),\mathrm{then}\frac{dy}{dx}$ is equal to ___________________.

Answer 1

We know

${\sec }^{-1}a={\cos }^{-1}\frac{1}{a}$

$\therefore {\sec }^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)={\cos }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$ .....(1)

Now,

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

$\Rightarrow y={\cos }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+{\sin }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$ [Using (1)]

$\Rightarrow y=\frac{\mathrm{\pi }}{2}\left({\sin }^{-1}a+{\cos }^{-1}a=\frac{\mathrm{\pi }}{2}\right)$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=0$

Thus, if $y={\sec }^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x-1}}\right)+{\sin }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$, then $\frac{dy}{dx}=0$.


If $y={\sec }^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x-1}}\right)+{\sin }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$ , then $\frac{dy}{dx}$ is equal to ___0___.