We're asked to determine the derivative
of the given function,
using the limit definition
of the derivative.
Limit definition of the
derivative is given here,
where f prime of x, or
the derivative of f of x,
equals a limit, as h approaches
zero, of this quotient.
And this quotient is called
the difference quotient.
So, using the given
function f of x, we'll have
f prime of x equals the
limit, as h approaches zero,
of, again, our difference quotient.
So, looking at the
numerator, we want to find
f of the quantity x plus h.
Notice x plus h is the
input into the function f,
so wherever we see an x, we'll substitute
the quantity x plus h.
So f of the quantity x plus
h is going to be three,
and instead of x squared, we have
the quantity x plus h squared
minus, instead of five x,
we'll have five times
the quantity x plus h,
and then plus seven.
So all of this is f of
the quantity x plus h.
And then we have minus f
of x, so we'll have minus.
And to make sure we subtract
the entire function,
we need the quantity three x
squared minus five x plus seven
in a set of parentheses.
Without the parentheses, we'd
only subtract the first term
of the function f of x,
not the entire function.
Notice our denominator is just h.
Notice how we cannot
perform direct substitution
to determine this limit, because
we'd have division by zero.
So our next step is going to
be to simplify the numerator.
To do this, let's begin by
squaring the quantity x plus h.
So f prime of x equals the
limit, as h approaches zero.
It is important to keep
writing this limit notation
to have proper notation.
So we're going to have three,
and then times the quantity.
When we square the quantity
x plus h, we'll have
x squared plus two h x plus h squared.
Let's go and take a moment and show
how to square the quantity x plus h.
There are no shortcuts here.
To square the quantity x plus
h, we'll have two factors
of x plus h, and then
we'll have four products.
One, two, three, four.
So we'll have x squared plus h x plus h x
and plus h squared.
Combining like terms, we do
get x squared plus two h x
plus h squared, which
we already have here.
And for this step, we'll leave
everything else the same.
So we have minus five
times the quantity x plus h
plus seven, minus the
quantity three x squared
minus five x plus seven.
And the denominator is still h.
So I think an important
part of doing this correctly
is just taking your time
and showing detailed work.
So for our next step, we'll
clear the parentheses.
So we'll distribute three here.
Because of the subtraction,
we'll distribute
negative five here, and
because of this subtraction,
we can think of
distributing a negative one.
So f prime of x equals the
limit, as h approaches zero,
of...
Distributing three, we'd
have three x squared
plus six h x, plus three h squared.
Distributing negative
five, we have minus five x
minus five h, still have plus seven.
Distributing negative one we'll
have minus three x squared
plus five x minus seven.
Again, denominator is still h.
We now will combine the
like terms in the numerator,
and several of the terms
should be opposites.
We have two x squared terms.
We have two x terms.
And we also have two constants.
And notice how three x
squared minus three x squared
is zero.
Negative five x plus five x is zero.
And seven minus seven is zero.
So the numerator simplifies
nicely to six h x
plus three h squared minus five h.
And the denominator is still h.
Let's continue on the next slide.
Notice how we still have the
issue of division by zero,
for which I had to perform
direct substitution,
but now notice in the numerator,
all the terms have a common factor of h,
so we'll factor h out of the numerator
and then simplify this quotient.
If we factor h out of the numerator,
we would have h times
times the quantity six x
plus three h minus five.
Notice here we have h over
h, which simplifies to one.
So now we just have the limit,
as h approaches zero, of the
quantity six x plus three h
minus five, and notice how the
only term affected by h now
is this term here, three h,
and as h approaches zero,
three h approaches zero,
and therefore this limit
is just equal to six x minus five.
And now we have our derivative function.
F prime of x equals the
quantity six x minus five.
Before we go, though,
I do want to go back up
to this step here, and
show an alternative method
for simplifying this quotient.
So going back to f prime
of x equals the limit,
as h approaches zero, of...
Again, this quotient here,
six h x, plus three h squared,
minus five h, all over h.
Because we're dividing by a
monomial, we could simply this
by dividing each term
in the numerator by h.
This limit is equal to the
limit, as h approaches zero,
of six h x divided by h, plus
three h squared divided by h,
minus five h divided by h.
And now if we simplify
each fraction individually,
we will get the same limit.
So now simplifying h
over h simplifies to one.
Here we have a common factor of h.
The denominator simplifies to one.
Here we just have h to
the first, and h over h
simplifies to one here as well.
So we have the limit,
as h approaches zero,
of, here we just have six x.
Here we have plus three h,
and here we have minus five.
Notice the function is now in
the same form as we see here,
using a different method for
simplifying our quotient.
As h approaches zero,
three h approaches zero,
giving us the same limit
of six x minus five,
which is our derivative function.
I hope you found this helpful.
