Welcome to a proof
of the product rule of differentiation.
The product rule tells us how to find the derivative
when we have a product of two functions.
As we see written here using function notation
and here with just f and g.
The derivative of a product of two functions
is the first function, f,
times the derivative of the second function,
which would be g prime,
plus the second function, which is g,
times the derivative of the first function,
which would be f prime.
So again, these two formulas are exactly the same.
One's in function notation and one's just using f and g.
To prove the product rule, we will go back to
the limit definition of a derivative.
So we should already be familiar with this limit
definition of a derivative where we have the limit
as h approaches zero of the difference quotient
which is the quantity f of x plus h,
minus f of x,
divided by h.
And this is the limit definition
of the derivative of function f of x.
So the limit definition of the product of function f and g,
notice how instead of just f of x plus h,
we have f of x plus h,
times g of x plus h.
Instead of just minus f of x,
we have minus f of x,
times g of x
and our denominator is still h.
Now the next step is a little bit tricky.
What we're going to do here in green is subtract
f of x plus h times g of x
and then add the same product.
So we have plus f of x plus h,
times g of x.
And we might ask, how do we know to subtract this product
and then to add the same product resulting in zero?
Unfortunately, I don't have an easy answer,
we just take quite a bit of thought and some creativity.
Now for the next step in the proof,
we're going to break this fraction into two separate fractions.
So looking at the numerator,
we're actually going to break the fraction here.
Looking at these two products on the left,
notice how they both contain a factor of f of x plus h.
So for the first fraction, we're going to factor that out
which is this factor here.
Notice how that would leave us with g of x plus h,
minus g of x,
which we have here, divided by h.
And then for the second fraction, notice how the two
terms on the right share a common factor of g of x.
So if we factor out g of x for the second fraction,
notice how we're left with f of x plus h,
minus f of x.
Again, over the common denominator of h.
Next, we take this limit of a sum and write it as
a sum of two limits, which we see here.
Then because these two limits involve a product,
we have a product here and a product here,
we can write these as the product of limits,
which we see here.
So we have the limit as h approaches zero
of f of x plus h,
times the limit of this difference quotient
involving function g,
plus the limit as h approaches zero of g of x,
times the limit as h approaches zero of
the difference quotient involving function f of x.
So now we actually have our proof.
Since h is approaching zero,
this is approaching zero.
So this first limit is equal to f of x,
which notice is our first function here,
times the limit as h approaches zero of this,
which is the difference quotient for function g of x.
So this limit is equal to the derivative of g of x.
So we have f of x times g prime of x,
plus the limit as h approaches zero of g of x,
which is just g of x or the second function,
times the limit as h approaches zero
of the difference quotient for f of x.
So this last limit is equal to the derivative of f of x.
Which is exactly what we were trying to prove.
We've just shown the derivative
of f times g is equal to f, the first function,
times the derivative of the second function,
which is g prime,
plus the second function, g,
times the derivative of the first function,
which is f prime.
So here's our proof of the product rule.
I realize most calculus textbooks have this proof,
but I hope this explanation still helps.