Welcome to a proof of the product rule
of differentiation which states that the derivative
of f of x times g of x with respect to x
equals f of x times g prime of x plus g of x
times f prime of x.
Or, we can say the derivative is equal
to the first function times the derivative
of the second function plus the second function
times the derivative of the first function.
So to begin, if we apply the limit definition
of the derivative shown here for our function
f of x times g of x, the derivative is going
to be equal to the limit as h approaches zero of,
now looking at the definition,
for f of the quantity x plus h,
notice for our function, when the input is
the quantity x plus h,
we get f of the quantity x plus h
times g of the quantity x plus h.
And then, for minus f of x,
for our function when the input is x,
we get f of x times g of x.
And now, for the next step,
we're going to form an equivalent fraction
in a different form by adding and subtracting
the same term in the numerator.
So for the next step,
we subtract f of the quantity x plus h
times g of x and then add f of the quantity
x plus h times g of x.
Notice how this sum would be zero.
So this is an equivalent fraction.
It's just in a different form.
Next we'll write this limit of a single fraction
as a limit of two fractions
where for the first fraction we'll use
the difference of these two terms
over the denominator of h.
And the second fraction will have the sum
of these two terms over the denominator of h.
But looking at this difference here,
notice how we do have a common factor
of f of the quantity x plus h.
So we'll also factor the first fraction at the same time.
Looking at these two terms, notice how
we have a common factor of g of x.
So again, we'll also factor the second fraction
at the same time.
So for our first limit, we have the limit
as h approaches zero of,
again, we factored out the common factor
of f of the quantity x plus h
from these first two terms.
If we want to check this, we can distribute here
to verify we get the same two terms we have here.
And then, for the limit as h approaches zero
of our second fraction, we can also distribute
g of x here to verify we get these two terms
in the original numerator.
And now we're going change the form
of these two fractions by writing them
as a product.
So looking at this first fraction,
notice how, because we have f of the quantity
x plus h times this difference in the numerator,
we can rewrite this as f of the quantity x plus h
times this fraction which is actually
the difference quotient for g of x.
Then looking at our second limit,
again, because we have g of x times this difference
in the numerator, we can write this fraction
as g of x times this difference quotient
which is the difference quotient for f of x.
Now notice how we have a limit of a product here
as well as a limit of a product here.
Whenever we have a limit of a product,
we can write that as a product of limits.
So for our next step, this limit becomes
the limit as h approaches zero of f
of the quantity x plus h times the limit
as h approaches zero of the difference quotient for g of x.
And then we still have plus.
And then here we have the limit
as h approaches zero of g of x times the limit
as h approaches zero of the difference quotient for f of x.
And now we have everything we need.
Let's take this line to the next slide.
If we evaluate each of these limits,
we should have the product rule.
Notice for this first limit,
as h approaches zero, this limit is going
to be equal to f of x.
Next we have the limit as h approaches zero of,
this is the difference quotient for g of x
which equals g prime of x
and then plus the limit as h approaches zero of g of x.
This is not affected by h, and therefore,
this limit is equal to g of x.
And then we have times the limit
as h approaches zero of,
this is the difference quotient for f of x
which is equal to f prime of x.
And therefore, we have our proof.
This is equal to f of x times g prime of x
plus g of x times f prime of x.
I hope you found this helpful.