Differentials and implicit differentiation Consider this example. Suppose $x$ is a function of two variables $s$ and $t$,
$$x = \sin(s+t)$$
Taking the differential as in doing implicit differentiation [1],
$$dx = \cos(s+t)(ds+dt) = \cos(s+t)dt + \cos(s+t)ds$$

I know the right way of taking the differential of $x$ is by
$dx = \dfrac{\partial x}{\partial s}ds + \dfrac{\partial x}{\partial t}dt$ [2]
But why do the above method [1], which I cannot make any sense of mathematically, give the same result as [2]?

I do not understand method [1] because $s$ and $t$ are supposed to be independent variables not functions to be differentiated. i.e. $ds$ just means $s-s_0$, same with $dt$.
Briefly, in implicit differentiation, I find it very helpful to treat every variable as a function.  Because in 
[1], the traditional method, chain rule the chain rule should apply to every variable or
[2], the differential method, one of ds, dx, dt, etc. should apply to every variable.

The exception is in the traditional, non differential method.  If you're taking derivative with respect to x, then x doesn't need to be a function.  Simply take derivative of x like normal.
This is a special case: you may consider a simple substitution $y = s + t$

$x = \sin(y)$

$dx = \cos(y)dy$

$dx = \cos(s+y)(ds + dt)$

if you do this with $x=s^t$

$y = s^t$

$x = y$

$dx = dy$

$dx = ts^{t-1}ds + s^t\ln(s)dt$

which isn't useful at all...