Answer
$$\frac{dy}{dx}=\ln2\ln3\ln4\times2^{3^{4^x}}\times3^{4^x}\times4^x$$
Work Step by Step
$$y=2^{3^{4^x}}$$ $$\frac{dy}{dx}=\frac{d(2^{3^{4^x}})}{dx}$$
Let $u=4^x$, then $y=2^{3^u}$. According to Chain Rule, we have $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{{d(2^{3^{u}})}}{du}\frac{d(4^x)}{dx}$$
Let $v=3^u$, then $y=2^v$. According to Chain Rule, we have $$\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{{d(2^{v})}}{dv}\frac{d(3^u)}{du}\frac{d(4^x)}{dx}$$ $$\frac{dy}{dx}=\ln2\times2^v\times\ln3\times3^u\times\ln4\times4^x$$ $$\frac{dy}{dx}=\ln2\ln3\ln4\times2^{3^{4^x}}\times3^{4^x}\times4^x$$
