Jon P. answered • 05/06/15

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I assume you mean "product," not quotient. Remember, a product is the result of a multiplication, but a quotient is the result of a division.

So what we want to prove is:

lim [(fg)(x + Δx) - (fg)(x)] / Δx = f'(x)g(x) + g'(x)f(x)

Δx->0

Let's work with the quantity on the left side and see what we get...

[(fg)(x + Δx) - (fg)(x)] / Δx

= [f(x + Δx) g(x + Δx) - f(x) g(x)] / Δx

Let's do a bit of a trick... Subtract and add f(x + Δx) g(x) in the middle of the numerator. We get:

[f(x + Δx) g(x + Δx) - f(x + Δx) g(x) + f(x + Δx) g(x) - f(x) g(x)] / Δx

That doesn't change anything because we added and subtracted the same thing.

Now break this into two separate expressions:

[f(x + Δx) g(x + Δx) - f(x + Δx) g(x)] / Δx + [f(x + Δx) g(x) - f(x) g(x)] / Δx

In the first expression, factor out f(x + Δx), and in the second expression, factor out g(x):

f(x + Δx) [g(x + Δx) - g(x)] / Δx + g(x) [f(x + Δx) - f(x)] / Δx

Now, remember that we're taking the limit if this whole expression, so what we're really trying to determine is:

lim (f(x + Δx) [g(x + Δx) - g(x)] / Δx + g(x) [f(x + Δx) - f(x)] / Δx)

Δx->0

= lim f(x + Δx) [g(x + Δx) - g(x)] / Δx

Δx->0

+ lim g(x) [f(x + Δx) - f(x)] / Δx

Δx->0

Since the limit of f(x + Δx) is just f(x), we can take that term out of the first limit and call it f(x). And we can do the same with g(x) in the second limit. So we get:

f(x) lim [g(x + Δx) - g(x)] / Δx + g(x) lim [f(x + Δx) - f(x)] / Δx

Δx->0 Δx->0

But look at the expressions INSIDE the limits now.

lim [g(x + Δx) - g(x)] / Δx is just g'(x).

Δx->0

And...

lim [f(x + Δx) - f(x)] / Δx is just f'(x)

Δx->0

So that gives us f(x)g'(x) + g(x)f'(x)

And that's exactly what we were attempting to prove!