Answer
$y'=\sinh^{-1}({\dfrac{x}{3}})$
Work Step by Step
given: $y=x\sinh^{-1}(\frac{x}{3})-\sqrt{9+x^2}$
$y'=\dfrac{d}{dx}(x\sinh^{-1}(\frac{x}{3})-\sqrt{9+x^2})$
or, $y'=\frac{d}{dx}x\sinh^{-1}(\frac{x}{3})-\frac{d}{dx}\sqrt{9+x^2}$
Apply product rule.
$y'=(1\times\sinh^{-1}(\frac{x}{3})+x\times\frac{d}{dx}\sinh^{-1}(\frac{x}{3})-\frac{d}{dx}\sqrt{9+x^2}$
Now, apply chain rule.
$y'=\sinh^{-1}({\frac{x}{3}})+(x)\dfrac{1}{\sqrt{1+(\frac{x}{3})^2}}
(\dfrac{1}{3})-\dfrac{1}{2\sqrt{9+x^2}} (2x)$
or, $=\sinh^{-1}({\frac{x}{3}})+\dfrac{x}{3\sqrt{\dfrac{x^2+9}{9}}}-\dfrac{x}{\sqrt{9+x^2}}$
or, $=\sinh^{-1}({\dfrac{x}{3}})+\dfrac{x}{\sqrt{{x^2+9}}}-\dfrac{x}{\sqrt{x^2+9}}$
Thus, $y'=\sinh^{-1}({\dfrac{x}{3}})$
