Proof: The Product Rule of Differentiation

This video explains the proof of the product rule using the limit definition of the derivative.

[ghostridersinthesky]

I love this proof, its broken down so much better than other videos Ive watched.

[qualquan]

Don't like the arbitrarily making f(x+h) = f(x) on the left but not doing same later on the right. Very inconsistent.
Easier if f(x+h) depicted as f(x)+df and g(x+h) as g(x)+dg
Then THEIR product is simply: f(x).g(x) +f(x).dg +g(x).df +df.dg
Then subtracting f(x).g(x) and dividing by dx gives:
f(x).dg/dx +g(x).df/dx +df.dg/dx
or f(x).g' + g(x).f' while the remaining term df.dg/dx is removed as it contains ALL the averaging error (not because it is zero)
BTW, since f(x+h) is f(x) + df
So f(x+h). g' = f(x).g' + df.g'
Then df.g' is eliminated as it contains averaging error and NOT because it is zero.

[ritzzzblitzz]

This part 1:33 was killing me earlier, now it is cleared. Thank you sir.

[thebobhannah]

Very well presented. Thanks for the useful explanation.

[AlbertEinsteinJr]

Great work Sir, this was very useful.