Solutions to Worksheet for Sections 2.6 and 2.7

Tangents, Velocity, and the Derivative

Math 1a

February 11, 2008

1. Let f(x) = x3. Use the definition of the derivative to find f ′(2).

Solution. We have

f ′(2) = limh→0

(2 + h)3 − 23

h3

= limh→0

(8 + 12h + 6h2 + h3)− 23

h3

= limh→0

12h + 6h2 + h3

h3

= limh→0

(12 + 6h2 + h3

)= 12.

2. Let f(x) =√

x.

(a) Use the definition of the derivative to find f ′(4).

Solution. We have

f(x)− f(4)x− 4

=√

x− 2x− 4

=√

x− 2x− 4

·√

x + 2√x + 2

=(x− 4)

(x− 4) (√

x + 2)

=1

(√

x + 2)

as x→ 4, this expression goes to f ′(4) = 1/4.

(b) Find f ′(x) and give its domain.

Solution. Do the same algebra with 4 above replaced by a, and you get f ′(a) =1

2√

a. In other

words, f ′(x) =1

2√

xfor all x. This assumes that x > 0 otherwise the limit does not exist.

(c) Is f differentiable at zero? Use a graph to illustrate why or why not.

Solution. Graphically, we can see that the tangent line to f at zero is vertical. Analytically, wecan see that the difference quotient at zero is

f(h)− f(0)h

=

√h

h=

1√h

which, as h → 0+ has no limit (the limit is ∞). This doesn’t even get into the fact that f isnot defined for x < 0, and we usually require a limit from both sides before a function can bedifferentiable.

Here’s a useful fact: for any numbers A and B:

A3 −B3 = (A−B)(A2 + AB + B2)

3. Let f(x) = 3√

x = x1/3. Use the definition of the derivative to find f ′(x) and give its domain.

Solution. The difference quotient for f ′(a) is

x1/3 − a1/3

x− a· x

1/3 + x1/3a1/3 + a2/3

x1/3 + x1/3a1/3 + a2/3=

���x− a

(���x− a)(x2/3 + x1/3a1/3 + a2/3

)as x→ a this tends to f ′(a) =

13a1/3

. So f ′(x) =13x−1/3 for all x > 0.

4. Repeat with f(x) = x2/3.

Solution. The difference quotient for f ′(a) is

x2/3 − a2/3

x− a=

(x1/3 − a1/3

) (x1/3 + a1/3

)x− a

=x1/3 + a1/3

x2/3 + x1/3a1/3 + a2/3→ 2a1/3

3a2/3

as x→ a. So f ′(a) =23a−1/3, or f ′(x) =

23x−1/3 for all x > 0.