Calculus I, Section 2.8, #58
The Derivative as a Function
(a) If g(x) = x2/3 , show that g 0 (0) does not exist.1
g(0 + h) g(0)
h!0
h
g(h) g(0)
= lim
h!0
h
2/3
h
02/3
= lim
h!0
h
h2/3
= lim
h!0 h
1
= lim 1/3
h!0 h
g 0 (0) = lim
1
Now, if h ! 0 , then h1/3 is negative-valued and h1/3
!
1
and h1/3 ! 1. Thus the limit as h ! 0 does not exist.
1. If h ! 0+ , then h1/3 is positive-valued
(Note that either of these one-sided limits becoming infinite is enough to conclude g 0 (0) does not exist.
More about this in a future lesson.)
(b) If a 6= 0, find g 0 (a).
g(a + h) g(a)
h
2/3
(a + h)
a2/3
= lim
h!0
h
g 0 (a) = lim
h!0
and we’re stuck. No more algebra is possible to simplify this. Let’s try the other definition of derivative.
g(x)
x
2/3
x
= lim
x!a
x
g 0 (a) = lim
x!a
= lim
x1/3
g(a)
a
a2/3
a
2
x
x!a
2
a1/3
a
From di↵erence of squares, we get
= lim
x!a
x1/3 + a1/3
x
x1/3
a
a1/3
Now we’ll write the denominator as a di↵erence of cubes, to look for a common factor with the
numerator.
= lim
x1/3 + a1/3
x!a
= lim
x!a
x1/3
x1/3
3
x1/3
a1/3
a1/3
3
x1/3 + a1/3 x1/3 a1/3
a1/3 x2/3 + x1/3 a1/3 + a2/3
x1/3 + a1/3
x!a x2/3 + x1/3 a1/3 + a2/3
2a1/3
= 2/3
3a
2
= 1/3
3a
= lim
1 Stewart,
Calculus, Early Transcendentals, p. 164, #58.
Calculus I
The Derivative as a Function
(c) Show that y = x2/3 has a vertical tangent line at (0,0).
Our work in part (a) with the one-sided limits shows that when a < 0, the slope of the tangent line
is negative and the tangent line becomes steeper with negative slope as we get closer to 0. Similarly,
when a > 0, the slope of the tangent line is positive and becomes the tangent line becomes steeper
with positive slope the close we get to 0.
Thus, there is a vertical tangent line at x = 0.
(d) Illustrate part (c) by by graphing y = x2/3 .
Using WolframAlpha, we get
which cleary shows the vertical tangent line at (0,0).