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# Product rule proof

Why does the product rule work?

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• how can we say that d/dx (f(x).g(x)) = lim(h --> 0) [f(x+h).g(x+h) - f(x).g(x)]/h
did not understand about the numerator of the limit or the derivative. •  A good, formal definition of a derivative is, given `f(x)` then `f′(x) = lim(h->0)[ (f(x-h)-f(x))/h ]` which is the same as saying if `y = f(x)` then `f′(x) = dy/dx`. `dy = f(x-h)-f(x)` and `dx = h`. Since we want h to be 0, `dy/dx = 0/0`, so you have to take the limit as h approaches 0.

So now, we're starting with a particular form of f(x) where `f(x) = g(x)*k(x)`. So, we plug g(x)*k(x) into our original equation to figure out its derivative.
`f′(x) = lim(h->0)[ (f(x-h)-f(x))/h ]`
`f′(x) = lim(h->0)[ (g(x-h)*k(x-h) - g(x)*k(x)) / h ]`
• The little trick he uses is not sufficient enough for me to prove to myself the product rule. Seems like a shot in the blind. Thankfully we already know what the product rule is so we sort of know what we need to add the expression but there has to be a more intuitive way to prove the product rule? Also he's multiplying the entire derivations of 2 functions so why isn't it ((f(x+h)-f(x))*(g(x+h)-g(x)))/h ?????? Am guessing because the product of 2 derived functions is not the same as derivation of products of functions? This would mean that f(x)*g(x) is now one new function, it's not 2 seperate functions but one as in multiplications where ab is not 2 terms but 1 term, so f(x)*g(x)=F(x), if we evaluate F(x) we get (F(x+h)-F(x))/h and like I said, F(x)=f(x)*g(x). So replace F(x) and F(x+h) with corresponding terms and you get what Sal got, did I get this right? I really do need to get this right otherwise my brain will refuse to memorize it. That is, it's easier for me to understand something instead memorizing the entire thing. That way I can always a problem down into what I know and therefore I can always find the formula for whatever problem I might have. Also watching Implicit differentiation and logarithmic differentiation has led me to believe this could be proven with natural logarithm but how? •  It is not a shot in the blind at all. It is a clever use of the fact that adding and subtracting the same value does not change an expression. The ability to see that such a device can work is the key to many proofs in mathematics, and it is this type of insight that separates the mathematicians from those that just do math without thinking about what it means. It is this type of insight and intuition, that being, the ability to leverage the rules of mathematics creatively that produces much of the beauty in math.

I think you do understand Sal's (AKA the most common) proof of the product rule.

Having said that, YES, you can use implicit and logarithmic differentiation to do an alternative proof:
y=f(x)g(x)
ln(y) = ln (f(x)g(x)) = ln(f(x)) + ln(g(x))
Take the derivative of both sides:
y'/y = f'(x)/f(x) + g'(x)/g(x)
Solve for y'
y' = y(f'(x)/f(x) + g'(x)/g(x))
Since y = f(x)g(x)
y' = f(x)g(x)(f'(x)/f(x) + g'(x)/g(x))
Multiply it all out and you get
y' = g(x)f'(x)+f(x)g'(x)
Since y' = d/dx[y] and y = f(x)g(x) we have
d/dx[f(x)g(x)] = g(x)f'(x) + f(x)g'(x).
• at , why is the limit of the product the product of the limits? Have any previous videos discussed this? • There are other ways to do it? This way seems pretty straight forward; how can you prove it any other way? • Let y=v.u, where v and u are functions in x
so y+δy=(v+δv)(u+δu) when we let x change by a small amount
if we expand the brackets we get
y+δy= v.u + u.δv + v.δu + δv.δu
as y=v.u,we can eliminate v.u from the right and y from the left
we then divide by δx to get δy/δx= u .δv/δx +v .δu/δx +(δv.δu)/δx
as δx tends to 0, (δv.δu)/δx becomes insignificant
so,we end up with dy/dx=u .dv/dx + v .du/dx,
which is another way of writing the product rule
• How is the `lim h->0 f(x+h)` the same as `f(x)`? that wouldnt be true for a discontinuous function. the `limit h->0 f(0+h)` is infinity but f(0) is not defined. So those expressions are not equal to each other. • the derivative is only defined where a function is defined. Going a bit farther you will see that the derivative (which is a limit) can only exist if the function is continuous at that point. If the function is not defined at a you certainly cannot take the derivative at x=a, since a isn't in the domain of the function and you cannot set up a different quotient at a.

We often say "differentiability implies continuity" since to have a derivative at a point a function must be continuous there (not just defined there). Even if it is continuous, it still might not be differentiable--think of the absolute value function at x=0. So continuity is a necessary but not sufficient condition for differentiability.
• At why doesnt he do FOIL? • at sal left a space and added and subtracted something to get in the required form but what about the real things which we get after multiplying {f(x+h)-f(x)/h}X {g(x+h)-g(x)/h} • from where did sal come up with the first equation written in white..........
i know i saw it somewhere at khan academy jst help me find that video • So if we can claim that the the limit as h→0 of f(x+h) is f(x), as was stated in the video at , Why can't you evaluate it as f(x+h)g(x+h)-f(x)g(x)/h = f(x)g(x+h)-f(x)g(x)/h = f(x)((g(x+h)-g(x))/h), which would be equal to f(x)g'(x). This result is clearly wrong, but I can't see where exactly I've made a mistake.  