Infinitely Nested Trig Function...'s derivative?

$\begin{aligned} y & = \sin(x+\tan(x+\sin(x+\tan(x+\sin(x+\tan(\cdots)))))) \\ & = \sin(x+\tan(x+y)) \\ \implies \frac {dy}{dx} & = \cos(x + \tan(x+y))\left(1+\sec^2(x+y)\left(1+\frac {dy}{dx}\right)\right) & \small \color{#3D99F6} \text{By chain rule} \\ \frac {dy}{dx}\bigg|_{x = \pi} & = \cos(\pi + \tan(\pi+{\color{#3D99F6}0}))\left(1+\sec^2(\pi+{\color{#3D99F6}0})\left(1+\frac {dy}{dx}\bigg|_{x = \pi}\right)\right) & \small \color{#3D99F6} \text{Note that }y(\pi) = 0 \\ & = (-1)\left(1+1+\frac {dy}{dx}\bigg|_{x = \pi}\right) \\ & = -2-\frac {dy}{dx}\bigg|_{x = \pi} \\ & = \boxed{-1} \end{aligned}$

Step 1: Sub expression into itself

$y=sin(x+tan(x+y))$

Step 2: Implicitly differentiate

$\frac{dy}{dx} = cos(x+tan(x+y)) (1 + sec^{2}(x+y)) (1+\frac{dy}{dx})$

Step 3: Rearrange for $\frac{dy}{dx}$

$\frac{dy}{dx} = -\frac{(sec^{2}(y+x)+1)(cos(tan(y+x)+x)}{(sec^{2}(y+x))(cos(tan(y+x)+x)-1}$

Step 4: $y = 0$ satisfies the original equation (Notice that y is in [-1,1] as sin can only produce between -1 and 1))

$0=sin(pi+tan(pi+0))$

Step 5: Plugin $x = \pi$ and $y = 0$

$\frac{dy}{dx} = -\frac{(sec^{2}(0+\pi)+1)(cos(tan(0+\pi)+\pi)}{(sec^{2}(0+\pi))(cos(tan(0+\pi)+\pi)-1}$

$\frac{dy}{dx} = -\frac{(1+1)(cos(0+\pi)}{(1)(cos(0+\pi)-1}$

$\frac{dy}{dx} = -\frac{(2)(-1)}{(1)(-1) -1}$

$\frac{dy}{dx} = -\frac{(-2)}{(-2)}$

$\frac{dy}{dx} = -1$

Done !