(1)f′(3)=limx→3f(x)−f(3)x−3=limx→3√x−√3x−3=limx→3√x−√3x−3√x+√3√x+√3=limx→3x−3(x−3)(√x+√3)=limx→31√x+√3=1√3+√3=12√3
(2)f′(3)=limh→0f(3+h)−f(3)h=limh→0√3+h−√3h=limh→0√3+h−√3h√3+h+√3√3+h+√3=limh→0(3+h)−3h(√3+h+√3)=limh→0hh(√3+h+√3)=limh→01√3+h+√3=limh→01√3+√3=12√3
We see that
the derivative of f at
x=3 is
12√3, i.e.
f′(3)=12√3.
Just as in the prior slide computing instantaneous velocity,
notice that in both computations (until the denominator was
cancelled) if we plugged
x=3 in, we would get the
indeterminate form
00 (
try
it) so we had to
do
more work.
Also, again notice that the work required in either computation
is similar.
Be sure you practice
method (2)!