We're asked to find the derivative of f of x
using the limit definition of the derivative shown below,
where we have the limit of the difference quotient
as h approaches 0,
which means in our case, f prime of x
is equal to the limit of,
again, the difference quotient as h approaches 0,
where the difference quotient
is f of the quantity x plus h minus f of x,
all divided by h.
In our case, f of the quantity x plus h
is equal to the square root
of the quantity 3 minus 2 times the quantity x plus h,
and then minus f of x is just minus the square root
of the quantity 3 minus 2x,
all divided by h.
Notice we cannot
find the limit by performing direct substitution
because if we tried, we would have division by 0.
So what we're going to do is rationalize the numerator
by multiplying the numerator and denominator
by the conjugate of the numerator.
And let's show the algebra below.
Starting with the difference quotient,
we'll multiply the numerator and denominator
by the square root of 3 minus 2 times
the quantity x plus h,
plus the square root of the quantity 3 minus 2x.
Multiplying the conjugates in the numerator,
the first product is the square root of
3 minus 2 times the quantity x plus h times itself,
which gives us 3 minus 2 times the quantity x plus h.
The next 2 products will be opposites and simplified as 0.
The last product is negative square root of the quantity
3 minus 2x times positive square root
of the quantity 3 minus 2x,
which gives us minus the quantity, 3 minus 2x.
In the denominator,
we have h times the sum of these two square roots.
And now we clear the parentheses in the numerator
and then simplify.
We have 3 minus 2x minus 2h,
and then minus 3 plus 2x,
all over the same denominator.
Combining like terms in the numerator,
we have 3 minus 3, as well as negative 2x plus 2x.
Notice the numerator simplifies to negative 2h.
We have negative 2h all over the same denominator.
Notice in this form,
we can see we have a common factor of h
between the numerator and denominator,
h divided by h simplifies to 1.
It's now going back up to divisional limit.
Divisional limit is equal to
the limit of this negative 2,
divided by the sum of
the square root of 3 minus 2 times the quantity x plus h,
and the square root of 3 minus 2x,
as h approaches 0.
Notice in this form can now find the limit
by performing direct substitution.
If we set h equal to 0, we just have negative 2
over these square root of the quantity, 3 minus 2x
plus these square root of the quantity, 3 minus 2x
which gives us 2 times the square root
of the quantity 3 minus 2x.
And we can simplify one more time.
Notice how there's a common factor of 2
between the numerator and denominator that simplify to 1,
which gives us f prime of x is equal to negative 1
divided by the square root of the quantity 3 minus 2x.
I hope you found this helpful.