Answer
$$\frac{d}{dx}(uvw)=wv\frac{du}{dx}+wu\frac{dv}{dx}+uv\frac{dw}{dx}$$
$$\frac{d}{dx}(u_1u_2u_3u_4)=u_2u_3u_4\frac{du_1}{dx}+u_1u_3u_4\frac{du_2}{dx}+u_1u_2u_4\frac{du_3}{dx}+u_1u_2u_3\frac{du_4}{dx}$$
$$\frac{d}{dx}(u_1u_2u_3...u_n)=u_2u_3u_4...u_n\frac{du_1}{dx}+u_1u_3u_4...u_n\frac{du_2}{dx}+u_1u_2u_4...u_n\frac{du_3}{dx}+...+u_1u_2u_3...u_{n-1}\frac{du_n}{dx}$$
Work Step by Step
a) Find $\frac{d}{dx}(uvw)$: $$\frac{d}{dx}(uvw)=\frac{d}{dx}(uv\times w)$$
Apply the Derivative Product Rule for 2 functions $uv$ and $w$:
$$\frac{d}{dx}(uvw)=\frac{d}{dx}(uv)\times w+\frac{dw}{dx}\times(uv)$$
Now, to find $\frac{d}{dx}(uv)$, we apply the Derivative Product Rule again for 2 functions $u$ and $v$:
$$\frac{d}{dx}(uvw)=w\Big(\frac{du}{dx}\times v+\frac{dv}{dx}\times u\Big)+uv\frac{dw}{dx}$$ $$\frac{d}{dx}(uvw)=wv\frac{du}{dx}+wu\frac{dv}{dx}+uv\frac{dw}{dx}$$
b) Find $\frac{d}{dx}(u_1u_2u_3u_4)$: $$\frac{d}{dx}(u_1u_2u_3u_4)=\frac{d}{dx}(u_1u_2u_3\times u_4)$$
Apply the Derivative Product Rule for 2 functions $u_1u_2u_3$ and $u_4$:
$$\frac{d}{dx}(u_1u_2u_3u_4)=\frac{d}{dx}(u_1u_2u_3)\times u_4+\frac{du_4}{dx}\times(u_1u_2u_3)$$
Now, to find $\frac{d}{dx}(u_1u_2u_3)$, we apply the result from part a):
$$\frac{d}{dx}(u_1u_2u_3u_4)=u_4\Big(u_2u_3\frac{du_1}{dx}+u_1u_3\frac{du_2}{dx}+u_1u_2\frac{du_3}{dx}\Big)+(u_1u_2u_3)\frac{du_4}{dx}$$ $$\frac{d}{dx}(u_1u_2u_3u_4)=u_2u_3u_4\frac{du_1}{dx}+u_1u_3u_4\frac{du_2}{dx}+u_1u_2u_4\frac{du_3}{dx}+u_1u_2u_3\frac{du_4}{dx}$$
c) Find $\frac{d}{dx}(u_1u_2u_3...u_n)$
Generalizing the results from part a) and part b), we can say that $$\frac{d}{dx}(u_1u_2u_3...u_n)=u_2u_3u_4...u_n\frac{du_1}{dx}+u_1u_3u_4...u_n\frac{du_2}{dx}+u_1u_2u_4...u_n\frac{du_3}{dx}+...+u_1u_2u_3...u_{n-1}\frac{du_n}{dx}$$
