Review for prelim 2:
(x)
• Definition of derivate: f 0 (x) = limh→0 f (x+h)−f
.
h
• The concept of tangent line, normal line and the slope of a curve (which is defined
as the slope of the tangent line).
• Left and right derivative.
• Differentiability implies continuity.
• Differentiation rules (in particular, the Chain rule, inverse rule, and the derivative
of trig & inverse trig functions).
• Differentiate f (x)g(x) (use f g = eg ln f ).
• Use chain rule in “related rates” problems.
• Ways to show derivatives exist & to calculate them: differentiation rules, or use
the definition of derivative as a limit.
dy
d
: see y as a function of x, do dx
• Implicit differentiation: f (x, y) = 0, calculate dx
on both sides and use the chain rule.
• Linearization based on derivative: when h is “small”, f (a + h) ≈ f (a) + hf 0 (a), or,
if we let x = a + h, f (x) ≈ f (a) + (x − a)f 0 (a).
• Definition of absolute min/max, local min/max, and critical points.
• Continuous function on finite closed interval must have minimum and maximum.
• Mean Value Theorem: be careful about the conditions (continuous on [a, b], differentiable on (a, b)).
• First derivative test for monotonity and local min/max. When looking for local
extrema of f :
Step 1: Decide the domain of f .
Step 2: List (1) points where f 0 = 0.
Step 3: List all points where f 0 does not exist, including the boundary points if
there are any.
Step 4: Use the first derivative to decide if f is increasing or decreasing between
those points.
Step 5: Decide which points are local max and which are local min.
Please make sure that you are familiar with the shape of the graph of some common
functions. Remember, sketching the graph of functions can often give you the correct intuition for solving problems.
Please remember that unless told otherwise, only theorems explicitly stated in our textbook can be cited without justification in exams.
Exercises:
1. ex sin y = x2 + y 4 , find
dy
dx .
1
2
Solution: Differentiate with respect to x on both sides, we have: (sin y+xy 0 cos y)ex sin y =
yex sin y −2x
2x + 4y 0 y 3 , so y 0 = 4ysin
3 −x cos yex sin y .
2. Find the domain, range, local maximum, local minimum, absolute maximum, absolute minimum, of f (x) = (tan x)tan x .
Solution: f is defined where tan x > 0, which implies kπ < x < kπ + π/2, where k is
any integer. So the domain of f is the union of all open intervals (kπ, kπ + π/2), where k
is an integer.
f 0 (x) = (etan x ln(tan x) )0 = (tan x ln(tan x))0 etan x ln(tan x) = (sec2 x ln(tan x)+sec2 x)etan x ln(tan x) =
sec2 xetan x ln(tan x) (ln(tan x) + 1). Because etan x ln(tan x) > 0, sec2 x > 0, f 0 (a) = 0 if and
only if ln(tan a) + 1 = 0, i.e. a ∈ {tan−1 1e + kπ}, where k is an integer. Furthermore, because ln and tan are both increasing functions, f 0 < 0 on (kπ, kπ + tan−1 1e ),
and f 0 > 0 on (kπ + tan−1 1e , kπ + π/2). Hence, by the first derivative test, f reaches
−1
its (local as well as absolute) minimum at tan−1 1e + kπ, and f (tan−1 1e + kπ) = e−e .
As x approaches π/2 from the left, tan x → ∞, so is (tan x)tan x , hence the range of f is
−1
[f (tan−1 1e + kπ), ∞) = [e−e , ∞). It has no (local or absolute) maximum.
3. Find the critical points of function g(x) = |(x − 1)3 (x + 1)4 | and the intervals on
which it is increasing or decreasing.
Solution: (x + 1)4 (
≥ 0, so (x − 1)3 (x + 1)4 is positive when x > 1 and non-positive when
(x − 1)3 (x + 1)4 ,
x>1
x ≤ 1, hence g(x) =
. By definition, since all real numbers are
3
4
−(x − 1) (x + 1) , x ≤ 1
in the interior of the domain of g, a point is a critical point if g is not differentiable at that
point or if g 0 = 0 at that point. When x > 1, g 0 (x) = 3(x − 1)2 (x + 1)4 + 4(x − 1)3 (x + 1)3 =
(x−1)2 (x+1)3 (3x+3+4x−4) = (x−1)2 (x+1)3 (7x−1) > 0, so there are no critical point on
(1, ∞). When x < 1, g 0 (x) = −3(x−1)2 (x+1)4 −4(x−1)3 (x+1)3 = −(x−1)2 (x+1)3 (7x−1),
so −1 and 1/7 are the only critical points on (−∞, 1). Finally, at 1, the left-hand derivative
and the right-hand derivative of g are both 0, so g 0 (1) = 0, 1 is also a critical point.
By the first derivative test one can show that g is decreasing on (−∞, −1], increasing
on [−1, 1/7], decreasing on [1/7, 1] and finally increasing on [1, ∞).
4. Suppose a spherical planet has a constant radius R. When flying at height h above
2h
dh
its surface, the visible region of the surface has area S = 2πR
R+h . Write dt as a function of
R, h and dS
dt .
3
Solution: Differentiate with respect to t, we have
dh
dt
=
(R+h)2
2πR3
dS
dt .
dS
dt
(R+h)−h
= 2πR2 dh
dt (R+h)2 =
2πR3 dh
.
(R+h)2 dt
So,