The basic reason is that one is simply composing the derivatives just as one composes the functions. Derivatives are linear approximations to functions. When you compose the functions, you compose the linear approximations---not a surprise.
I'm going to try to expand on Harry Gindi's answer, because that was the only way I could grok it, but in somewhat simpler terms.
The way to think of a derivative in multiple variables is as a linear approximation. In particular, let $f: R^m \to R^n$ and $q=f(p)$. Then near $p$, we can write $f$ as $q$ basically something linear plus some "noise" which "doesn't matter" (i.e. is little oh of the distance to $p$). Call this linear map $L: R^m \to R^n$.
Now, suppose $g: R^n \to R^s$ is some map and $r = g(q)$. We can approximate $g$ near $q$ by $r$ plus some linear map $N$ plus some "garbage" which is, again, small.
For simplicity, I'm going to assume that $p,q,r$ are all zero. This is ok, because one can just move one's origin around a bit.
So, as before, applying $f$ to a point near zero corresponds loosely to applying the linear transformation $L$. Applying $g$ to a point near zero corresponds loosely to applying $N$. Hence applying $g \circ f$ corresponds up to some ignorable "garbage" to the map $N \circ L$.
This means that $N \circ L$ is the linear approximation to $g \circ f$ at zero, in particular this composition is the derivative of $g \circ f$.
Hint:
$$\frac{d}{d \theta}\sin(\theta^{\circ}) = \frac{d}{d\theta} \sin\left(\frac{\pi}{180^{\circ}} \cdot \theta^{\circ}\right).$$
Now, can you apply the chain rule?
Best Answer
Consider $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}f(u)}{\mathrm{d}u}\dfrac{\mathrm{d}u}{\mathrm{d}x}$. Take the second derivative (use the chain rule, product rule, and chain rule, respectively, in that order): $$\dfrac{\mathrm{d}^2y}{\mathrm{d}^2x}=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}f(u)}{\mathrm{d}u}\dfrac{\mathrm{d}u}{\mathrm{d}x}\right)=\dfrac{\mathrm{d}f(u)}{\mathrm{d}u}\dfrac{\mathrm{d}^2u}{\mathrm{d}x^2}+\dfrac{\mathrm{d}^2f(u)}{\mathrm{d}u^2}\left(\dfrac{\mathrm{d}u}{\mathrm{d}x}\right)^2$$