# 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 $dt$.

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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...

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Shouldn't that be $dx = ts^{t-1}ds + s^t\ln(s)dt$? Which is of the form $dx = \dfrac{\partial x}{\partial s}ds + \dfrac{\partial x}{\partial t}dt$ –  Ron Mar 12 '13 at 11:25
of cause. (fixing) –  V-X Mar 13 '13 at 6:59