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Derivatives
Definition and Notation
If ( )
y = f x then the derivative is defined to be ( )
( ) ( )
0
lim
h
f x h f x
f x
h
→
If ( )
y = f x then all of the following are
equivalent notations for the derivative.
( ) ( ( )) ( )
df dy d
f x y f x Df x
dx dx dx
If ( )
y = f x all of the following are equivalent
notations for derivative evaluated at x = a.
( ) ( )
x a
x a x a
df dy
f a y Df a
dx dx
=
= =
Interpretation of the Derivative
If ( )
y = f x then,
( )
m f a
= is the slope of the tangent
line to ( )
y = f x at x = a and the
equation of the tangent line at x = a is
given by ( ) ( )( )
y f a f a x a
( )
f a
is the instantaneous rate of
change of ( )
f x at x = a.
- If ( )
f x is the position of an object at
time x then ( )
f a
is the velocity of
the object at x = a.
Basic Properties and Formulas
If ( )
f x and ( )
g x are differentiable functions (the derivative exists), c and n are any real numbers,
( ) ( )
c f c f x
- ( f g ) f ( x ) g ( x )
- ( f g ) f g f g
= + – Product Rule
2
f f g f g
g g
- Quotient Rule
( )
d
c
dx
( )
n n 1
d
x n x
dx
−
= – Power Rule
( ( )) ( )
( ( )) ( )
d
f g x f g x g x
dx
This is the Chain Rule
Common Derivatives
( )
d
x
dx
( )
sin cos
d
x x
dx
( )
cos sin
d
x x
dx
( )
2
tan sec
d
x x
dx
( )
sec sec tan
d
x x x
dx
( )
csc csc cot
d
x x x
dx
( )
2
cot csc
d
x x
dx
( )
1
2
sin
d
x
dx
x
−
( )
1
2
cos
d
x
dx
x
−
( )
1
2
tan
d
x
dx x
−
( )
( )
ln
x x
d
a a a
dx
( )
x x
d
dx
e = e
( ( ))
ln , 0
d
x x
dx x
( )
ln , 0
d
x x
dx x
( ( ))
log , 0
ln
a
d
x x
dx x a
Chain Rule Variants
The chain rule applied to some specific functions.
d n n 1
f x n f x f x
dx
−
( )
f x f ( x )
d
f x
dx
e = e
ln
f x d
f x
dx f x
sin cos
d
f x f x f x
dx
cos sin
d
f x f x f x
dx
2
tan sec
d
f x f x f x
dx
( [ ]) [ ] [ ]
sec f ( ) x f ( ) x sec f ( ) x tan f ( ) x
d
dx
′
1
2
tan
f x d
f x
dx
f x
−
Higher Order Derivatives
The Second Derivative is denoted as
( )
2
2
2
d f
f x f x
dx
′′ = = and is defined as
f x f x
′′ = ′ , i.e. the derivative of the
first derivative,
f ′ x.
The n
th
Derivative is denoted as
( )
n
n
n
d f
f x
dx
= and is defined as
( )
( )
n n 1
f x f x
−
= , i.e. the derivative of
the ( n -1)
st
derivative,
( )
n 1
f x
−
Implicit Differentiation
Find y
if
2 9 3 2
sin 11
x y
x y y x
−
e + = +. Remember
y = y x here, so products/quotients of x and y
will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to
differentiate as normal and every time you differentiate a y you tack on a y
(from the chain rule).
After differentiating solve for y ′.
2 9 2 2 3
2 9 2 2
2 9 2 9 2 2 3
3 2 9
3 2 9 2 9 2 2
2 9 3 2 cos 11
2 9 3 2 cos 11
2 9 cos
2 9 cos 11 2 3
x y
x y
x y x y
x y
x y x y
y x y x y y y y
x y
y x y x y y y y y
x y y
x y y y x y
−
−
− −
−
− −
e
e
e e
e
e e
Increasing/Decreasing – Concave Up/Concave Down
Critical Points
x = c is a critical point of
f x provided either
f ′ c = 0 or 2.
f ′ c doesn’t exist.
Increasing/Decreasing
- If
f x 0
> for all x in an interval I then
f x is increasing on the interval I.
- If
f x 0
< for all x in an interval I then
f x is decreasing on the interval I.
- If
f x 0
= for all x in an interval I then
f x is constant on the interval I.
Concave Up/Concave Down
- If
f ′′ x > 0 for all x in an interval I then
f x is concave up on the interval I.
- If
f ′′ x < 0 for all x in an interval I then
f x is concave down on the interval I.
Inflection Points
x = c is a inflection point of
f x if the
concavity changes at x = c.
Related Rates
Sketch picture and identify known/unknown quantities. Write down equation relating quantities
and differentiate with respect to t using implicit differentiation ( i.e. add on a derivative every time
you differentiate a function of t ). Plug in known quantities and solve for the unknown quantity.
Ex. A 15 foot ladder is resting against a wall.
The bottom is initially 10 ft away and is being
pushed towards the wall at
1
4
ft/sec. How fast
is the top moving after 12 sec?
x
is negative because x is decreasing. Using
Pythagorean Theorem and differentiating,
2 2 2
x y 15 2 x x 2 y y 0
After 12 sec we have ( )
1
4
x = 10 − 12 = 7 and
so
2 2
y = 15 − 7 = 176
. Plug in and solve
for y
( )
1
4
7 176 0 ft/sec
y y
Ex. Two people are 50 ft apart when one
starts walking north. The angle θ changes at
0.01 rad/min. At what rate is the distance
between them changing when θ = 0.5rad?
We have θ ′ = 0.01rad/min. and want to find
x
. We can use various trig fcns but easiest is,
sec sec tan
x x
We know θ = 0.5so plug in θ ′ and solve.
( ) ( )( )
sec 0.5 tan 0.5 0.
0.3112 ft/sec
x
x
Remember to have calculator in radians!
Optimization
Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for
one of the two variables and plug into first equation. Find critical points of equation in range of
variables and verify that they are min/max as needed.
Ex. We’re enclosing a rectangular field with
500 ft of fence material and one side of the
field is a building. Determine dimensions that
will maximize the enclosed area.
Maximize A = xy subject to constraint of
x + 2 y = 500. Solve constraint for x and plug
into area.
( )
2
A y y
x y
y y
Differentiate and find critical point(s).
A 500 4 y y 125
By 2
nd
deriv. test this is a rel. max. and so is
the answer we’re after. Finally, find x.
( )
x = 500 − 2 125 = 250
The dimensions are then 250 x 125.
Ex. Determine point(s) on
2
y = x + 1 that are
closest to (0,2).
Minimize ( ) ( )
2 2
2
f = d = x − 0 + y − 2 and the
constraint is
2
y = x + 1. Solve constraint for
2
x and plug into the function.
( )
( )
2
2 2
2
2
x y f x y
y y y y
Differentiate and find critical point(s).
3
2
f ′ = 2 y − 3 ⇒ y =
By the 2
nd
derivative test this is a rel. min. and
so all we need to do is find x value(s).
2 3 1 1
2 2
2
x = − 1 = ⇒ x = ±
The 2 points are then
( )
1 3
2
2
, and
( )
1 3
2
2
