Lesson08: Derivatives (worksheet Solutions)

Solutions to Worksheet for Sections 2.6 and 2.7 Tangents, Velocity, and the Derivative Math 1a October 12, 2007 1.

Let f (x) = x3 . Use the definition of the derivative to find f 0 (2).

Solution. We have (2 + h)3 − 23 h→0 h3 (8 + 12h + 6h2 + h3 ) − 23 = lim h→0 h3 2 12h + 6h + h3 = lim h→0 h3
= lim 12 + 6h2 + h3 = 12.

f 0 (2) = lim

h→0

2.

Let f (x) =

√

x.

(a) Use the definition of the derivative to find f 0 (4). Solution. We have √ √ √ f (x) − f (4) x−2 x−2 x+2 = = ·√ x−4 x−4 x−4 x+2 (x − 4) √ = (x − 4) ( x + 2) 1 = √ ( x + 2) as x → 4, this expression goes to f 0 (4) = 1/4. (b) Find f 0 (x) and give its domain. 1 Solution. Do the same algebra with 4 above replaced by a, and you get f 0 (a) = √ . In other 2 a 1 0 words, f (x) = √ for all x. This assumes that x > 0 otherwise the limit does not exist. 2 x

(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, we can see that the difference quotient at zero is √ f (h) − f (0) h 1 = =√ h h h which, as h → 0+ has no limit (the limit is ∞). This doesn’t even get into the fact that f is not defined for x < 0, and we usually require a limit from both sides before a function can be differentiable. Here’s a useful fact: for any numbers A and B: A3 − B 3 = (A − B)(A2 + AB + B 2 ) 3.

Let f (x) =

√ 3

x = x1/3 . Use the definition of the derivative to find f 0 (x) and give its domain.

Solution. The difference quotient for f 0 (a) is  x1/3 − a1/3 x1/3 + x1/3 a1/3 + a2/3 x− a 
· 1/3 = 1/3 1/3 2/3  2/3 1/3  x−a x +x a +a ( x − a) x + x a1/3 + a2/3 as x → a this tends to f 0 (a) = 4.

1 1 . So f 0 (x) = x−1/3 for all x > 0. 3 3a1/3

Repeat with f (x) = x2/3 .

Solution. The difference quotient for f 0 (a) is

x1/3 − a1/3 x1/3 + a1/3 x1/3 + a1/3 x2/3 − a2/3 2a1/3 = = 2/3 → 2/3 1/3 1/3 2/3 x−a x−a x +x a +a 3a as x → a. So f 0 (a) =

2 −1/3 2 a , or f 0 (x) = x−1/3 for all x > 0. 3 3