If $f\left(x\right)={\sin }^{-1}\sqrt{\frac{x-\beta }{\alpha -\beta }}\text{ and }g\left(x\right)={\mathrm{Tan}}^{-1}\sqrt{\frac{x-\beta }{\alpha -x}}$ then show that   $f^1\left(x\right)=g^1\left(x\right)\cdot (\beta <x<\alpha )$

Let $f\left(x\right)={\sin }^{-1}\sqrt{\frac{x-\beta }{\alpha -\beta }}$

$f^1\left(x\right)=\frac{1}{\sqrt{1-\frac{x-\beta }{\alpha -\beta }}}\cdot \frac{d}{dx}\sqrt{\frac{x-\beta }{\alpha -\beta }}\left(\because \frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}\right)$

$=\frac{1}{\frac{\sqrt{\alpha -x}}{\sqrt{\alpha -\beta }}}\frac{1}{\sqrt{\alpha -\beta }}\frac{d\left(\sqrt{x-\beta }\right)}{dx}$

$=\frac{1}{\frac{\sqrt{\alpha -x}}{\sqrt{\alpha -\beta }}}\frac{1}{\sqrt{\alpha -\beta }}\frac{1}{2\sqrt{x-\beta }}$
$\therefore f^{\mathrm{\prime }}\left(x\right)=\frac{1}{2\sqrt{(\alpha -x)(x-\beta )}}$.....(1)

And  $g\left(x\right)={\mathrm{Tan}}^{-1}\sqrt{\frac{x-\beta }{\alpha -x}}$$\boxed{\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}}$

$g^1\left(x\right)=\frac{1}{1+\frac{x-\beta }{\alpha -x}}\cdot \frac{d}{dx}\sqrt{\frac{x-\beta }{\alpha -x}}$

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

$=\frac{\sqrt{\alpha -x}(\alpha -x)}{2(\alpha -\beta )\sqrt{x-\beta }}\cdot \frac{\alpha -\beta }{(\alpha -x{)}^2}=\frac{1}{2\sqrt{(\alpha -x)(x-\beta )}}$.....(2)

from (1) & (2)  $\therefore f^1\left(x\right)=g^1\left(x\right)$