Download What is a derivative?

The derivative as the
slope of the tangent line
(at a point)
Index
FAQ



Video help:
MIT!!!
http://ocw.mit.edu/courses/mathematics/
18-01sc-single-variable-calculus-fall2010/1.-differentiation/
Index
FAQ
What is a derivative?
A
function, which gives the:
 the rate of change of a
function in general
 the slope of the line tangent
to the curve in general
Index
FAQ
What is a differential quotient?
Just a number!
 the rate of change of a function at
a given point
 the slope of the line tangent to the
curve at a certain point
 The substitutional value of the
derivative

Index
FAQ
The tangent line
single point
of intersection
Index
FAQ
slope of a secant line
f(a) - f(x)
a-x
f(x)
f(a)
Index
x
a
FAQ
slope of a (closer) secant line
f(a) - f(x)
a-x
f(x)
f(a)
Index
x
x
a
FAQ
closer and closer…
Index
a
FAQ
watch the slope...
Index
FAQ
watch what x does...
Index
x
a
FAQ
The slope of the secant line gets
closer and closer to the slope of the
tangent line...
Index
FAQ
As the values of x get closer and
closer to a!
Index
x
a
FAQ
The slope of the secant lines
gets closer
to the slope of the tangent line...
...as the values of x
get closer to a
Translates to….
Index
FAQ
Differential quotient
lim
x
a
f(x) - f(a)
x-a
as x goes to a
Equation for the slope
Which gives us the the exact slope
of the line tangent to the curve at a!
Index
FAQ
Differential quotient: other form
f(x+h) - f(x)
lim
= f(x+h) - f(x)
h0 (x+h) - x
h
f(a+h)
h
f(a)
Index
a+h
a
(For this particular curve, h is a negative value)
FAQ
Velocity and other Rates of
Change
Rates of Change:
f  x  h  f  x
Average rate of change =
h
f  x  h  f  x
Instantaneous rate of change = f   x   lim
h 0
h
These definitions are true for any function.
Index
FAQ
Velocity and other Rates of Changephysical menaing of the differential
quotient
Consider a graph of displacement (distance traveled) vs. time.
B
distance
(miles)
A
Average velocity can be found by
taking:
change in position
s

s
change in time
t
t
time (hours)
Vave
f  t  t   f  t 
s


t
t
The speedometer in your car does not measure average
velocity, but instantaneous velocity.
f  t  t   f  t 
ds
 lim
t 0
dt
t
Index
V t  
(The velocity at one
moment in time.)
FAQ
Velocity and other rates of
change
Velocity is the first derivative of position.
Acceleration is the second derivative
of position.
Index
FAQ
Velocity
Example:
Free Fall Equation
1 2
s g t
2
s  16 t
2
1
s   32 t 2
2
ds
V
 32 t
dt
Speed is the absolute value of velocity.
Index
Gravitational
Constants:
ft
g  32
sec 2
m
g  9.8
sec 2
cm
g  980
sec2
FAQ
3.4 Velocity and other Rates of
Change
Acceleration is the derivative of velocity.
dv
a
dt
d 2s
 2
dt
If distance is in:
example:
a  32
feet
Velocity would be in:
feet
sec
Acceleration would be in:
Index
v  32t
ft
sec  ft
2
sec
sec
FAQ
Velocity and other Rates of
Change
acc neg
vel pos &
decreasing
acc neg
vel neg &
decreasing
acc zero
vel pos &
constant
acc zero
vel neg &
constant
acc pos
vel neg &
increasing
distance
acc pos
vel pos &
increasing
velocity
zero
acc zero,
velocity zero
time
Index
FAQ
Differentiability
To be differentiable, a function must be continuous
and smooth.
Derivatives will fail to exist at:
f
 x 
2
f  x  x3
x
corner
f
 x 
3
x
f
vertical tangent
Index
cusp
1, x  0
 1, x  0
 x  
discontinuity
FAQ
Theorem : f is differentiable on the interval (a,b).  f is continuous
on the interval (a,b).
Proof: Assume that f ’(c) exists for any c in (a,b).
f ’(c) = lim f(c+h) -f(c)
h 0
h
Then lim [ f(c+h)- f(c)]
.
h0
= f ’(c) • lim h
.
/•h
= f ’(c) • 0 = 0
h 0
.
, and from here we get lim f(c+h) = f(c) .
So lim [ f(c+h) - f(c)] = 0
.
.
h0
h0
So f is continuous at c for every c in (a,b).
Index
FAQ
Remark
The reverse of this theorem is not
true.
Example: Since the derivative of f(x)= 5x2+x+1 is f ’(x) = 10x+1,
which exists for every real number x. So f(x)= 5x2+x+1 is continuous
everywhere.
Counter example: We know that f(x) = |x| is continuous on R , but at
x=0 it’s not differentiable since:
lim l0+hl –l0l = lim lhl
, which approaches to
h 0
h
. h 0 h
Index
+1 if h 0
–1 if h0
FAQ
Differentiability
To be differentiable, a function must be continuous
and smooth.
Derivatives will fail to exist at:
2
f
 x 
f  x  x3
x
corner
f
 x 
3
f
x
vertical tangent
Index
cusp
1, x  0
 1, x  0
 x  
discontinuity
FAQ
Derivatives of some elementary
functions
If the derivative of a function is its slope, then for a
constant function, the derivative must be zero. There is
no change...
d
c  0
dx
example:
y 3
y  0
The derivative of a constant is zero.
Index
FAQ
Derivatives of some elementary
functions
We saw that if
y  x2 , y  2 x
.
This is part of a pattern.
d n
n 1
x

nx


dx
examples:
f  x  x
y  x8
4
f   x   4x
3
y  8 x 7
power rule
Index
FAQ
Rules for Differentiation
Find the horizontal tangents of:
y  x4  2x2  2
dy
 4 x3  4 x
dx
Horizontal tangents occur when slope = zero.
4 x3  4 x  0
x3  x  0
x  x 2  1  0
x  x  1 x  1  0
x  0, 1, 1
Index
Substituting the x values into the
original equation, we get:
y  2, y  1, y  1
(The function is even, so we
only get two horizontal
tangents.)
FAQ
Velocity and other Rates of
Change
Rates of Change:
f  x  h  f  x
Average rate of change =
h
f  x  h  f  x
Instantaneous rate of change = f   x   lim
h 0
h
These definitions are true for any function.
( x does not have to represent time. )
Index
FAQ
Derivatives of Trigonometric
Functions
Consider the function y  sin  
We could make a graph of the slope:



Now we connect the dots!
The resulting curve is a cosine curve.
Index
d
sin  x   cos x
dx

2
slope
1
0
0

2
1

1
0
FAQ
Derivatives of Trigonometric
Functions
Proof
sin( x  h )  sin x
h 0
h
(sin x )'  lim
sin x cos h  sin h cos x  sin x
 lim
h 0
h
sin x (cos h  1)  sin h cos x
 lim
h 0
h
sin x (cos h  1)
sin h cos x
 lim
 lim
h 0
h 0
h
h
d
sin x (cos h  1)
sin h cos x
sin x  lim
 lim
h

0
h0
dx
h
h
Index
FAQ
Derivative of the cosine
Function
Find the derivative of cos x: (cos x )'  lim
h 0
cos x (cos h  1)
sin h sin x
lim
 lim
h 0
h 0
h
h
cos( x  h )  cos x
h
cos x (cos h  1)  sin h sin x
h 0
h
cos x (cos h  1)
sin h sin x
 lim
 lim
h 0
h 0
h
h
 lim
Index
FAQ
Derivative of the cosine function
is sine (cont.)
cos x cos h  sin h sin x  cos x
 lim

h 0
h
cos x (cos h  1)  sin h sin x
 lim

h 0
h
cos x (cos h  1) sin h sin x
 lim


h 0
h
h
(cos h  1)
sin h
 cosx lim
 sin x lim

h 0
h

0
h
h
 cos x.0  sin x.1  sin x
Index
FAQ
Derivatives of Trigonometric
Functions
We can find the derivative of tangent x by using the
quotient rule.
2
2
d
tan x
dx
d sin x
dx cos x
cos x  cos x  sin x    sin x 
cos 2 x
Index
cos x  sin x
cos2 x
1
cos 2 x
sec 2 x
d
tan  x   sec 2 x
dx
FAQ
Derivatives of Trigonometric
Functions
Derivatives of the remaining trig functions
can be determined the same way.
d
sin x  cos x
dx
d
cot x   csc 2 x
dx
d
cos x   sin x
dx
d
sec x  sec x  tan x
dx
d
tan x  sec 2 x
dx
d
csc x   csc x  cot x
dx
Index
FAQ
The Derivatives of the Sum, Difference,
Product and Quotient
If u  x  and v  x  are derivable, and C is any constant,
then so is u  x   v  x  , u  x  v  x  , Cu  x  , and
u  x
. Its derivative is given by the formula
v  x

(1) u( x)  v( x)  u( x)  v( x)

(2) u( x )  v( x )  u( x )v( x )  u( x )v( x )

(3) Cu( x )  Cu( x )
Index

 u( x )  u( x )v( x )  u( x )v( x )
(4) 

( v( x)  0)
2

v ( x)
 v( x ) 
FAQ
Proof

(1) u( x)  v( x)  u( x)  v( x)
(1) Let y  u( x )  v( x ), we have to examine
y
u( x  x )  v ( x  x )  u( x )  v ( x )
lim
 lim
x 0 x
x 0
x
u( x  x )  u( x )  v( x  x )  v( x )

 lim
x 0
Index
x
FAQ

(1) u( x)  v( x)  u( x)  v( x)
u v
 lim (
 )  u( x )  v( x )
x 0 x
x
Thus u( x )  v( x ) is derivable and

u( x )  v( x )  u( x )  v( x )
A similar argument applies to u( x )  v ( x ),
that is

u( x )  v( x )  u( x )  v( x )
Index
FAQ

(2) u( x )  v( x )  u( x )v( x )  u( x )v( x )
(2) Let y  u( x )v ( x ), then we express y in terms
of u and v. Finally, we determine y   x  by
y
examining lim
x 0 x
y
u( x  x )v ( x  x )  u( x )v ( x )
y  lim
 lim
x 0 x
x 0
x
[u( x )  u ][v ( x )  v ]  u( x )v ( x )
 lim
x  0
x
Index
FAQ
u( x ) v  v ( x ) u  uv
 lim
x 0
x
 u( x )v( x )  u( x )v( x )
u
v
v
 lim[ v( x )  u( x )
 u ]
x 0 x
x
x
Thus, u( x )v( x ) is derivable and

u( x)v( x)  u( x)v( x)  u( x)v( x)
Index
FAQ

(3) Cu( x )  Cu( x )


HOMEWORK!!

Index
FAQ

 u( x )  u( x )v( x )  u( x )v( x )
(4) 

( v( x)  0)
2

v ( x)
 v( x ) 
u( x  x ) u( x )

y
v( x  x ) v( x )
y  lim
 lim
x 0 x
x0
x
u( x )  u u( x )

v( x )  v v( x )
 lim
x 0
x
Index
FAQ

 u( x )  u( x )v( x )  u( x )v( x )
(4) 

( v( x)  0)
2

v ( x)
 v( x ) 
u( x  x ) u( x )

y
v( x  x ) v( x )
y  lim
 lim
x 0 x
x0
x
u( x )  u u( x )

v( x )  v v( x )
 lim
x 0
x
Index
FAQ
[u( x )  u ]v( x )  u( x )[v( x )  v ]
 lim
x 0
[v( x )  v ]v( x )x
u
v
v( x ) 
u( x )
uv( x )  u( x )v
 lim

x

x

lim
x 0 [v ( x )  v ]v ( x ) x
x 0 [v ( x )  v ]v ( x )
Index
FAQ
dy dy du


dx du dx
Chain Rule
Chain Rule:
If f g is the composite of y  f  u  and u  g  x  ,
then:
f
example:
g   f at u  g  x   gat x  f ' ( g ( x))  g ' ( x)
f  x   sin x
g  x   x2  4
f   x   cos x
f   0  g   2
g  x   2x
cos  0   2  2
Index
Find:
f
g  at x  2
g  2  4  4  0
1 4
4
FAQ
Remark

f(g(x))’= f ’(g(x)) g’(x) says that to get the
derivative of the “nested functions” you multiply
the derivative of each one starting from left to
right and so on
Index
FAQ
Example for using Chain rule
Example : Find y’(1) for y = (3x2-2)3( 5x3-x-3)4
y ’= 3(3x2 -2)2 (3x2-2)’( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 ( 5x3-x-3)’
y ’= 3(3x2 -2)2 (6x)
( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 (15x2-1)
y ’(1) = 3(3-2)2 (6)
(5-1-3)4
+ (3-2)3 4 (5-1-3)3 (15-1)
= 74
YOUR TURN!
dy .
. 2 x - 1 , find
when x=1
dx
For y =
√5x2+4 .
Index
FAQ
Example for using Chain rule
y  sin  x 2  4 
d 2
y  cos  x  4    x  4 
dx
2
y  cos  x 2  4   2 x
Index
FAQ
Example for using Chain rule
d
cos 2  3 x 
dx
d
2 cos  3x    cos  3x 
dx
d
2 cos  3x    sin  3 x    3 x 
dx
2cos  3x   sin 3x   3
6cos  3x  sin 3x 
Index
2
d
cos  3 x  
dx
The chain rule can be used
more than once.
(That’s what makes the
“chain” in the “chain rule”!)
FAQ
Implicit Differentiation
x2  y 2  1
d 2 d 2 d
x 
y  1
dx
dx
dx
dy
2x  2 y
0
dx
dy
2y
 2 x
dx
Index
This is not a function,
but it would still be
nice to be able to find
the slope.
Do the same thing to both sides.
Note use of chain rule.
dy 2 x

dx 2 y
dy
x

dx
y
FAQ
Implicit Differentiation
This can’t be solved for y.
2 y  x 2  sin y
dy
2x
d
d 2 d

2y 
x  sin y
dx 2  cos y
dx
dx
dx
dy
dy
2  2 x  cos y
This technique is called
dx
dx
implicit differentiation.
dy
dy
2  cos y
 2x
1 Differentiate both sides w.r.t. x.
dx
dx
2 Solve for y’
dy
 2  cos y   2 x
dx
Index
FAQ
Implicit Differentiation
Implicit Differentiation Process
1. Differentiate both sides of the equation with respect
to x.
2. Collect the terms with y’=dy/dx on one side of the
equation.
3. Factor out y’=dy/dx .
4. Solve for y’=dy/dx .
Index
FAQ
Implicit Differentiation
Find the equations of the lines tangent and normal to the
2
2
x

xy

y
7
curve
x  xy  y  7
2
at (1, 2) .
2
Note product rule.
dy y  2 x

dx 2 y  x
dy
 dy

2x   x  y  2 y
0
dx
 dx

dy
dy
22
4
2x  x  y  2 y
0
2  2  1


m
dx
dx
4 1
5
2

2


1


dy
 2 y  x  y  2x
dx
Index
FAQ
Implicit Differentiation
Find the equations of the lines tangent and normal to the
curve
x  xy  y  7 at
4
m
5
Index
2
2
tangent:
4
y  2   x  1
5
4
4
y2 x
5
5
4
14
y  x
5
5
(1, 2) .
normal:
5
y  2    x  1
4
5
5
y2  x
4
4
5
3
y  x
4
4
FAQ
Implicit Differentiation
d2y
3
2
Find
if
2
x

3
y
7 .
2
dx
2 x3  3 y 2  7
6 x 2  6 y y  0
6 y y  6 x 2
6 x 2
y 
6 y
x2
y 
y
Index
y  2 x  x 2 y
y 
y2
2x x 2
y 
 2 y
y y
Substitute
2
2
2x x x
back into they
y 
 2
equation.
y y y
2x x 4
y 
 3
y y
FAQ
Derivatives of Inverse
Trigonometric Functions
We can use implicit
differentiation to find:
y  sin 1 x
sin y  x
dy
cos y
1
dx
Index
d
sin 1 x
dx
1.5
y  sin 1 x
1
y  sin x
0.5
d
d
sin y 
x
dx
dx
dy
1

dx cos y
-1.5
-1
-0.5
0
0.5
1
1.5
-0.5
-1
-1.5
FAQ
Derivatives of Inverse
Trigonometric Functions
d
sin 1 x
dx
d
d
sin y 
x
dx
dx
dy
1

dx
1  sin 2 y
We can use implicit
differentiation to find:
y  sin 1 x
sin y  x
dy
cos y
1
dx
dy
1

dx cos y
Index
dy
1

dx
1  x2
sin 2 y  cos2 y  1
cos2 y  1  sin 2 y
cos y   1  sin 2 y
But 

 y

2
2
so cos y is positive.
 cos y  1  sin 2 y
FAQ
Derivatives of Inverse
Trigonometric Functions
y  sin 1 x
sin y  x
dy
cos y
1
dx
dy
1

dx cos y
Index
dy
1

dx cos(sin 1 x)
dy
1

dx
1  x2
FAQ
Derivatives of Inverse
Trigonometric Functions
d
1
tan
x
Find
dx
y  tan 1 x
tan y  x
dy
sec y
1
dx
dy
1

dx sec 2 y
dy
1

dx sec 2 (tan 1 x)
2
Index
dy
1

dx 1  x 2
FAQ
Derivatives of Exponential and Logarithmic
Functions
Look at the graph of
ye
The slope at x = 0 appears to
be 1.
x
3
2
If we assume this to be
true, then:
1
lim
e
h 0
-3
-2
-1
0
-1
Index
1
x
2
0 h
e
1
h
0
3
definition of derivative
FAQ
Derivatives of Exponential and Logarithmic
Functions
Now we attempt to find a general formula for the
x
derivative of y  e using the definition.
d x
e xh  e x
e  lim
h 0
dx
h
 
h

e
1 
x
 e  lim 

h 0
 h 
e x  eh  e x
 lim
h 0
h
This is the slope at x = 0, which
we have assumed to be 1.
 x eh  1 
x
x
 lim  e 


e

1

e
h 0
h 

Index
FAQ
Derivatives of Exponential and
Logarithmic Functions
e
x
is its own derivative!
If we incorporate the chain rule:
d u
u du
e e
dx
dx
We can now use this formula to find the derivative of
Index
ax
FAQ
Derivatives of Exponential and
Logarithmic Functions
 
d x
a
dx
d ln a x
e
dx
d x ln a
e
dx
d
x ln a
e   x ln a 
dx
 

Index

d x
x
a  a ln a
dx
Incorporating the chain rule:
 
d u
du
u
a  a ln a
dx
dx
FAQ
Derivatives of Exponential and
Logarithmic Functions
So far today we have:
d u
u du
e e
dx
dx
 
d
du
u
u
a  a ln a
dx
dx
Now it is relatively easy to find the derivative of ln x .
Index
FAQ
Derivatives of Exponential and
Logarithmic Functions
y  ln x
e x
y
 
d y
d
e   x
dx
dx
y dy
e
1
dx
Index
dy 1
 y
dx e
d
1
ln x 
dx
x
d
1 du
ln u 
dx
u dx
FAQ
Derivatives of Exponential and
Logarithmic Functions
To find the derivative of a common log function, you
could just use the change of base rule for logs:
d
d ln x
1 d
1 1
log x 

ln x 

dx
dx ln10
ln10 dx
ln10 x
The formula for the derivative of a log of any base
other than e is:
d
1 du
log a u 
dx
u ln a dx
Index
FAQ
Derivatives of Exponential and
Logarithmic Functions
 
d u
u du
e e
dx
dx
d
du
u
u
a  a ln a
dx
dx
d
1 du
ln u 
dx
u dx
d
1 du
log a u 
dx
u ln a dx
Index
FAQ
Derivatives of Exponential and Logarithmic
Functions
Logarithmic differentiation
Used when the variable is in the base and the exponent
y = xx
ln y = ln
xx
ln y = x ln x
1 dy
1
 x   ln x
y dx
 x
Index
dy
 y 1 ln x 
dx
dy
 x x 1 ln x 
dx
FAQ