Download LECTURE 9

MTH 104
Lecture # 9
Calculus and Analytical
Geometry
Chain rule
Consider
d
 x  1
dx
100
2

let y   x  1
100
2
then y  u
dy
 100u
du
and u  x  1
2
100
99
du
 2x
dx
and
dy dy du


dx du dx
 100u  2 x
99
 100( x  1)  2 x
2
 200 x ( x  1)
2
multiply rates
Chain rule
If g is differentiable at x and f is differentiable at g(x) then
the composition fog is differentiable at x. Moreover, if
y  f ( g ( x)) and u  g ( x)
dy dy du
 
dx du dx
Then y=f(u) and
Alternatively
d
 f  g  x    f  g   x   f  g  x g  x 
dx
Derivative of outside
function
Derivative of
inside function
Example
Find
dy
if y  cos( x )
3
Let
ux
3
Then y  cosu
dy

du
 sin u
du

dx
3x
And
Rates of
change
multiply
2
dy dy du
 
dx du dx
 ( sin u)  (3x )
2
 3x sin x
2
3
Example
d
d
d
tan x  tan x  (2 tan x) tan x  2 tan x sec x
dx
dx
dx
2
2
2
Derivative of
outside function
Derivative of
inside function
d
1
d
d
 x 1  x  1  x  1 x  1
dx
2
dx
dx
1
x

 2x 
2 x 1
x 1
2
2
2
1
2
2
2
1
2
2
More examples
1) f ( x)  (3x  5x )
2 7
2) f (t ) 
Solution 1)
du
2
Let u  3x  5 x 
 3  10 x
dx
dy
7
Then y u 
 7u 6
du
dy dy du
 
dx du dx
dy
 7u 6  (3  10 x)
dx
 7(3x  5 x 2 )6 (3  10 x)
5
7t  9 2
2) f (t ) 
Let
5
7t  9 2
du
u  7t  9   7
dt
5
y
 5u  2  dy  10u  3
du
u2
dy dy du
 
dt du dt
dy
 10u  3  7
dx
dy
 70(7t  9)  3
dt
Generalized derivative formulas
d
du
 f (u )  f (u )
dx
dx
where u  g ( x)
Some examples are:
d  r
du
r

1
u  ru

dx  
dx
d
cos u    sin u du
dx
dx
d
du
2
cot u    csc u
dx
dx
d
csc u    csc u cot u du
dx
dx
d
du
sin u   cos u
dx
dx
d
tan u   sec 2 u du
dx
dx
d
du
sec u   sec u tan u
dx
dx
Example
d
d
sin(2 x)  cos 2 x dx  2 x   2cos 2x
dx
1.
2.
d 
1
 d

x

2
x

3


dx  x  2 x  3  dx
2 d
 x3  2 x  3
   x3  2 x  3

 dx 

1
3
3

   x  2 x  3 3x  2
2
3
2
3x  2

 x  2 x  3
2
3
2
d y
dx
2
Example Find
if y  sin(3 x )
2
2
Solution
dy d
 sin(3x )
dx dx
d
 cos(3x ) 3x
dx
2
2
2

 6 x cos(3x )
d y d
  6 x cos(3x )
dx dx
d
d
 6 x  cos  3x   cos(3x ) 6 x 
dx
dx
2
2
2
2
2
2
d
d
 6 x  cos  3x   cos(3x ) 6 x 
dx
dx
d

 6cos(3x )
 6 x   sin(3x )  3x   
dx


2
2
2
2
2
 6 x  6 x sin(3x )  6cos  3x
2
d y
 36 x sin  3x   6cos  3x
dx
2
2
2
2
2

2

Example Differentiate


y  ln x 2  1
1
y  2
(2 x)
x 1
2x
y  2
x 1



y  ln x 2  1
Use the Chain
Rule

x
Example Differentiate y 
x
ln x
y
Use the Quotient
ln x
Rule
1
 
ln x(1)  x 
x

y
2
ln x 
ln x  1
y
2
ln x 
Related rates
Consider a water is draining out of a conical filter. The
volume V, the height h and the radius r are all functions
of the elapsed time t.
Volume formula:
V

3
r 2h
dV
rate of change of V 
dt
dV  d

[ r h]
dt 3 dt
2
dV 
dr
2 dh
 [2hr  r
]
dt
3
dt
dt
The rate of change of V is related to the rates of
change both r and h
Ralated
rates
problem
Example Suppose that x and y are differentiable functions of t
3
and are related by the equation y  x . Find dy/dt at time
t=1 if x=2 and dx/dt=4 at time t=1.
solution
dx
Known: x  2,
dt
3
yx
4
Unknown:
t=1
Differentiating both sides with respect to t
dy d 3
2 dx


  x   3x
dt dt
dt
dy
dt
 3(2)
t 1
2
dx
dt
 12 4  48
t 1
dy
dt
t=1
Example Suppose x and y are both differentiable functions
of t and are related by the equation y = x2 + 3. Find dy/dt,
given that dx/dt = 2 when x = 1
Solution
Given dx/dt = 2 when x = 1
y = x2 + 3
d
d 2
[ y ]  [ x  3]
dt
dt
dy
dx
 2x
dt
dt
dy
dt
x 1
dx
 2(1)
dt
2 24
x 1
To find dy/dt
Procedure for solving related rates problems
Step 1. Assign letters to all quantities that vary with time and any
others that seem relevant to the problem. Give a definition
for each letter.
Step 2. Identify the rates of change that are known and the rates
of change that is to be found. Interpret each rate as a
derivative.
Step 3. Find an equation that relates the variables whose rates of
change were identified in Step 2. To do this, it will often be
helpful to draw an appropriately labeled figure that
illsutrates the relationship.
Step 4. Differentiate both sides of the equation obtained in Step 3
with respect to time to produce a relationship between the
known rates and the unknown rates of change.
Step 5. After completing Step 4, substitute all known values for
the rates of change and the variables, and then solve for
the unknown rate of change.
Example Suppose that z  x3 y 2 , where both x and y are
changing with time. At a certain instant when x=1 and y=2, x is
decreasing at the rate of 2 units/s, and y is increasing at the
rate of 3 units/s. How fast is z changing at this instant? Is z
increasing or decreasing?
solution
dx
Known:
dt
dy
dt
 2,
x 1
y 2
3
x 1
y 2
zx y
3 2
Differentiating with respect to t
dz d 3 2
  x y 
dt dt
dz
Unknown:
dt
x 1
y 2
dz
2 d
3
3 d
2




y
x

x
y



dt
dt
dt
dx
3 dy
 3x y
 2x y
dt
dt
2
dz
dt
dz
dt
2
 3(1) (2)
2
x 1
y 2
2
dx
dt
dy
 2(1) (2)
dt
3
x 1
y 2
 12(2)  4(3)  12 units/s
x 1
y 2
Negative sign shows that it is decreasing
x 1
y 2
Example A stone is dropped into a still pond sends out a circular
ripple whose radius increases at a constant rate of 3ft/s. How
rapidly is the area of enclosed by the ripple increases at the end
of 10 s?
Solution
let r= radius of circular ripple, A= Area enclosed by the
ripple
dA
dr
Given
 3, To find
dt
dt
r 10
Since radius is increasing with a constant rate of 3
ft/s, so after 10 s the radius will be 30 ft.
We know that
A   r2
dA
dr
 2 r
dt
dt
dA
dt
 2 (30)(3)  180 ft 2 / s
r 10