How to prove the product rule using differentials? Proving the chain rule ( for single variable functions) using differentials is rather direct. But it seems more difficult to prove the product rule ( or the quotient rule).
Let $y= f(x) \space \space  z= g(x)$ and $p= h(x)=f(x) g(x)=y\times z$.
Knowing in advance the product rule, I also know that the result to be reached is :
$ \frac {dp}{dx} = y'z + yz'$  ( using $y'$ and $z'$ for the derivatives  w.r.t. $x$ of $f$ and $g$).
Working backwards, I get :
$dp = y' z dx+ yz'dx$ ( multiplying both sides by $dx$ )
$ dp = y'dxz + yz'dx$ ( changing the order of factors)
$dp = dy z + y dz  $   ( with, by definition,  $dy = y'dx $ and $ dz = z'dx$).
But I cannot go further in order to relate this to the starting point I want to reach through  this backwards process , namely :
$ dp = p' dx = [yz]' dx$.
 A: $$d(\log(f(x)g(x))=d(\log(f(x)+\log(g(x))=d(\log(f(x))+d(\log(g(x)).$$
Then by the chain rule,
$$\frac{d(f(x)g(x))}{f(x)g(x)}=\frac{df(x)}{f(x)}+\frac{dg(x)}{g(x)}$$ or $$(f(x)g(x))'=g(x)f'(x)+f(x)g'(x).$$
A: In Bouasse, Cours de mathématiques générales ( p.19) , the proof of the product rule is given as an instantiation of a more general rule, namely :
If $y= f( x, z, t...)$ then $dy = \frac {\partial y}  {\partial x} dx + \frac {\partial y}  {\partial z} dz + \frac {\partial y}  {\partial t} dt$.
The application to a function $y = f(x, z) = x\times z$ gives :
$dy = \frac {\partial y}  {\partial x} dx + \frac {\partial y}  {\partial z} dz$.
But, as notes the author : $\frac {\partial y}  {\partial x} = z$ and $ \frac {\partial y}  {\partial z}= x$.
This allows to write :
$dy = zdx + xdz$
Which means that, in case both $x$ and $z$ are functions of some variable $t$, one can get, dividing both sides by $dt$
$ y'(t) = \frac {dy} {dt} = z \frac {dx} {dt} + x \frac {dz} {dt} = z \space  x' + x\space z' $
as desired.
