A question is given in my book to differentiate the following function w.r.t. $x$. $$y= \sin^{-1}\left(\frac{3x + 4\sqrt{1 -x^2}}{5}\right)$$ I did it as follows: $$ y = \sin^{-1}\left(\frac{3}{5}x + \frac{4}5\sqrt{1 -x^2}\right)$$ Making the substitutions $\frac 35 = \sin\alpha$ and $x = \cos\beta$ so that $\frac45 = \cos\alpha$ and $\sqrt{1-x^2} = \sin\beta$. $$\begin{aligned}\implies y & =\sin^{-1}\left(\sin\alpha \cos\beta + \cos\alpha\sin\beta\right)\\ \implies y &=\sin^{-1}(\sin(\alpha+\beta))\\ \implies y &=(\alpha+\beta)\\ \implies y &=\sin^{-1}\left(\frac35\right) + \cos^{-1}(x)\end{aligned}$$ which can be differentiated effortlessly.
My question is that since $3$,$4$ and $5$ are forming pythagorean triplets, the function is easily differentiable. But what if the function is modified as: $$\boxed{y= \sin^{-1}\left(\frac{2x + 4\sqrt{1 -x^2}}{5}\right)}$$ What's the appropriate method to differentiate this? (I'm interested in ways without chain rule.)
It's not true that $\arcsin\left(\frac{3x+4\sqrt{1-x^{2}}}{5}\right) = \arcsin\left(\frac{3}{5}\right)+\arccos\left(x\right)$. The mistake is that
$$\sin^{-1}(\sin(\alpha+\beta)) = (\alpha+\beta)$$
because it only works if $(\alpha+\beta) \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. In this case, $(\alpha+\beta) \in \left[\arcsin\left(\frac{3}{5}\right), \arcsin\left(\frac{3}{5}\right) + \pi\right]$.
But to answer your question, I really don't think there's a way to differentiate that function without the chain rule. I tried finding other representations of $\arcsin\left(\frac{2x+4\sqrt{1-x^{2}}}{5}\right)$, but even the complex definition of it has two logs containing a mess. The only way I can come up with is by either computing the definition of the derivative of that function or implicitly differentiating it by first applying $\sin$ on both sides (yet implicit differentiation is essentially the chain rule).