3.6 Chain rule
When gear A makes x turns, gear B makes u turns and gear C makes y turns.,
u turns 3 times as fast as x
y turns ½ as fast as u
dy 1

du 2
du
3
dx
So y turns 3/2
as fast as x
dy dy du

dx du dx
Rates are multiplied
The Chain Rule for composite functions
If y = f(u) and u = g(x) then y = f(g(x)) and
dy dy du

dx du dx
multiply rates
dy d
  f ( g ( x)   f ( g ( x)) g ( x)
dx dx
multiply rates
Find the derivative (solutions to follow)
4) f ( x)  sin(2 x)
1) f ( x)  (3x  5x2 )7
2) f ( x)  ( x  1)
3
3) f (t ) 
2
7
(2t  3) 2
2
5) f ( x)  tan( x2  1)
Solutions
1) f ( x)  (3x  5x2 )7
u  3x  5 x 2 
du
 3  10 x
dx
dy
y u 
 7u 6
du
7
2) f ( x)  3 ( x 2  1)2
u  x2  1 
y
2
 u3

dy
 7(3x  2 x 2 )6 (3  10 x)
dx
dy 2
 u
dx 3
du
 2x
dx
dy 2
 u
du 3
dy dy du

dx du dx
dy
 7u 6 (3  10 x)
dx

1
3

1
3
(2 x)
dy 2 2
 ( x  1)
dx 3

1
3
(2 x)
dy
4x

1
dx
3( x 2  1) 3
Solutions
3) f (t ) 
7
(2t  3) 2
 7(2t  3) 2
du
2
dx
dy
2
y  7u 
 14u 3
du
u  2t  3 
dy dy du

dx du dx
dy
 14u 3 2
dx
dy
28
3
 14(2t  3) (2) 
dx
(2t  3)3
4) f ( x)  sin(2 x)
du
u  2x 
2
dx
dy
y  sin(u ) 
 cos(u )
du
5) f ( x)  tan( x2  1)
du
2
u  x 1 
 2x
dx
dy
y  tan(u ) 
 sec2 (u )
du
dy
 2cos(u )  2cos(2 x)
dx
dy
 sec2 (u )(2 x)  2 x sec2 ( x 2  1)
dx
Outside/Inside method of chain rule
dy d
  f ( g ( x)   f ( g ( x)) g ( x)
dx dx
outside
inside
think of g(x) = u
derivative
of outside
wrt inside
derivative
of inside
Outside/Inside method of chain rule example
outside
d  2
 3x  x  1
dx 



1
3


f ( g ( x)) g ( x)
inside
d  2
 3x  x  1
dx 


derivative
of inside
derivative
of outside
wrt inside

1
3


1
3x 2  x  1
 3


2
3
 6 x  1

6x  1

3 3x 2  x  1
2
3
Outside/Inside method of chain rule
outside


d
d
3
3
sin    sin    f ( g ( x)) g ( x)
dx
dx
inside
derivative
of outside
wrt inside
d
3
2 d
 sin    3 sin    sin  
dx
dx
 3sin 2  cos
derivative
of inside
Outside/Inside method of chain rule


outside
inside
d
csc( 2  3)  f ( g ( x)) g ( x)
dx

derivative
of outside
wrt inside

d
csc( 2  3)   csc( 2  3)cot( 2  3)2
dx
2 csc( 2  3)cot( 2  3)
derivative
of inside
More derivatives with the chain rule
f ( x)  x
1 x
product
2
f ( x)  x 2
f ( x ) 
f ( x) 
1 



1
1  x2
2
 x3
1  x 2 
 x3
1
2

d
1  x2
dx
f ( x )  x 2
1
2

f ( x)  x 2 1  x 2
2


 1 x
1
x2 2
2

 1  x2
1
2

d
x2
dx

( 2 x)  1  x 2
1
2
1
2 2
f ( x ) 

1
2
2x
Simplify terms
2x
1
x2 2
1   2 x 1  



1  x 
1
x2 2

1
2

Combine with
common denominator

 x3  (1  x 2 )2 x
1
x2 2
1  
3 x 3  2 x

1  x2

1
2

 x3  2 x  2 x3
1
x2 2
1  
1
2
More derivatives with the chain rule
 3x  1 
f ( x)   2

 x 3
3
 3x  1 
f ( x )  3  2

 x 3
2
d  3x  1 


dx  x 2  3 
Quotient
rule

2
 3 x  1   ( x  3)3  (3 x  1)(2 x)
3 2
 
2
2
 x 3 
x 3

2



2
2
 3 x  1   (3 x  9)  (6 x  2 x)
3 2
 
2
 x 3 
x2  3

2


2
2
 3x  1   3x  9  6 x  2 x
3 2
 
2
 x 3 
x2  3

2














3(3 x  1) 2 ( 3 x 2  2 x  9)


4

x2  3



Radians Versus Degrees
1 

180
radians
The formulas for derivatives assume x is in radian measure.
sin (x°) oscillates only /180 times as often as sin (x)
oscillates. Its maximum slope is /180.
d/dx[sin (x)] = cos (x)
d/dx [sin (x°) ] = /180 cos (x°)
3.7 Implicit Differentiation
Although we can not
solve explicitly for y,
we can assume that y
is some function of x
and use implicit
differentiation to find
the slope of the curve
at a given point
y=f (x)
If y is a function of x then its derivative is
dy
dx
y2 is a function of y, which in turn is a function of x.
using the chain rule:
d  2
dy
y  2y


dx
dx
Find the
following
derivatives
wrt
x
1
d  2
y 
dx  
 
1
y
2
d
sin y 
dx
dy
cos y
dx
d  2 3
x y


dx

1
2
dy
dx
dy
x 3y
 y3 2 x
dx
2
2
Use product rule
Implicit Differentiation
y  x  sin( xy)
2
2
1. Differentiate both sides of the equation with respect to x,
treating y as a function of x. This requires the chain rule.
2y
dy
 dy

 2 x  cos( xy )  x  y (1) 
dx
 dx

2. Collect terms with dy/dx on one side of the equation.
dy
dy
 2 x  cos( xy)( x )  cos( xy) y
dx
dx
dy
dy
2 y  cos( xy)( x )  2 x  cos( xy) y
dx
dx
2y
3. Factor dy/dx
dy
(2 y  x cos( xy))  2 x  y cos( xy)
dx
4. Solve for dy/dx
dy 2 x  y cos( xy)

dx 2 y  x cos( xy)
Find equations for the
tangent and normal to the curve at (2, 4).
Use Implicit Differentiation
find the slope of the tangent at (2,4)
find the slope of the normal at (2,4)
Solution
x3  y3  9 xy  0
1. Differentiate both sides of the equation with respect to x,
treating y as a function of x. This requires the chain rule.
3x 2  3 y 2
dy
dy
 (9 x  y 9)  0
dx
dx
dy
dy
3x  3 y
 9 x  9 y)  0
dx
dx
2
2
2. Collect terms with dy/dx on one side of the equation.
dy
dy
3y
 9 x  9 y  3x 2
dx
dx
dy
(3 y 2  9 x)  9 y  3x 2
Factor dy/dx dx
2
3.
4.
dy 9 y  3x 2
 2
Solve for dy/dx
dx 3 y  9 x)
2
9(4)  3(2)
24 4
mtan 


2
3(4)  9(2) 30 5
mnormal
5

4
Find dy/dx
x  y  (x  y )
2
2 2
1. Write the equation of the tangent line at (0,1)
2. Write the equation of the normal line at (0,1)
Solution
x  y  ( x 2  y 2 )2
1. Differentiate both sides of the equation with respect to x,
treating y as a function of x. This requires the chain rule.
dy
dy
 2( x 2  y 2 )(2 x  2 y )
dx
dx
dy
dy
dy
1
 4 x3  4 x 2 y  4 xy 2  4 y 3
dx
dx
dx
1
2. Collect terms with dy/dx on one side of the equation.
3.
dy
dy
dy
 4 x2 y  4 y3
 1  4 x3  4 xy 2
dx
dx
dx
Factor dy/dx dy (1  4 x 2 y  4 y 3 )  1  4 x3  4 xy 2
dx
4. Solve for dy/dx
dy 1  4 x3  4 xy 2

dx 1  4 x 2 y  4 y 3
Find dy/dx
x  y  (x  y )
2
2 2
dy 1  4 x3  4 xy 2

dx 1  4 x 2 y  4 y 3
1. Write the equation of the tangent line at (0,1)
1
y  1  ( x  0)
3
or
y
1
x 1
3
2. Write the equation of the normal line at (0,1)
y  1  3( x  0)
or
y  3x  1
3.8 Higher Derivatives
The derivative of a function f(x) is a function
itself f ´(x). It has a derivative, called the
second derivative f ´´(x)
If the function f(t) is a position function, the
first derivative f ´(t) is a velocity function and
the second derivative f ´´(t) is acceleration.
d2y
f ( x)  2
dx
The second derivative has a derivative (the third derivative) and
the third derivative has a derivative etc.
d3y
f ( x)  3
dx
4
d
y
(4)
f ( x)  4
dx
n
d
y
(n)
f ( x)  n
dx
Find the second derivative for
x
f ( x) 
x 1
( x  1)(1)  x(1)
1
f ( x) 

2
( x  1)
( x  1) 2
f ( x) 
d  1  d
2

(
x

1)



2 
dx  ( x  1)  dx
2
f ( x)  2( x  1) (1) 
( x  1)3
3
Find the third derivative for
x
f ( x) 
x 1
In algebra we study relationships among variables
•The volume of a sphere is related to its radius
•The sides of a right triangle are related by
Pythagorean Theorem
•The angles in a right triangle are related to
the sides.
In calculus we study relationships between the rates of
change of variables.
How is the rate of change of the radius of a
sphere related to the rate of change of the
volume of that sphere?
Examples of rates-assume all variables are
implicit functions of t = time
Rate of change in radius of a sphere
dr
dt
Rate of change in volume of a sphere
dV
dt
Rate of change in length labeled x
dx
dt
Rate of change in area of a triangle
dA
dt
Rate of change in angle,
3.9

d
dt
Solving Related Rates equations
1. Read the problem at least three times.
2. Identify all the given quantities and the quantities
to be found (these are usually rates.)
3. Draw a sketch and label, using unknowns when
necessary.
4. Write an equation (formula) that relates the
variables.
5. ***Assume all variables are functions of time and
differentiate wrt time using the chain rule. The
result is called the related rates equation.
6. Substitute the known values into the related rates
equation and solve for the unknown rate.
Figure 2.43:
The balloon in Example 3.
Related
Rates
A hot-air balloon rising straight up from a level field is tracked
by a range finder 500 ft from the liftoff point. The angle of
elevation is increasing at the rate of 0.14 rad/min. How fast is the
balloon rising when the angle of elevation is is /4?
Given:
x  500 ft
d
 0.14 rad / min
dt
y

Find:
dy

when  
dt
4
x  500
Figure 2.43:
The balloon in Example 3.
Related
Rates
A hot-air balloon rising straight up from a level field is tracked
by a range finder 500 ft from the liftoff point. At the moment the
range finder’s elevation angle is /4, the angle is increasing at
the rate of 0.14 rad/min. How fast is the balloon rising at that
moment?
y
tan 
500
d
d y
tan  
dt
dt 500
1 dy
2 d
sec 

dt 500 dt

1 dy
sec2 ( )(.14) 
4
500 dt
1
1 dy
(.14) 
500 dt
2 
cos ( )
4
y

x
dy
 140 ft / sec
dx
Figure 2.44:
Figure for Example 4.
Related
Rates
A police cruiser, approaching a right angled intersection from
the north is chasing a speeding car that has turned the corner
and is now moving straight east. The cruiser is moving at 60
mph and the police determine with radar that the distance
between them is increasing at 20 mph. When the cruiser is .6
mi. north of the intersection and the car is .8 mi to the east,
what is the speed of the car?
ds
Given:
Find:
 20 mph
dt
dy
 60mph
dt
dx
when x  .8, y  .6
dt
Figure
2.44:
Figure for Example 4.
2
2
2
s x y
ds
dx
dy
2s  2 x  2 y
dt
dt
dt
dx
2(1)(20)  2(.8)  2(.6)(60)
dt
dx
 70mph
dt
Given:
ds
 20 mph
dt
dy
 60mph
dt
Find:
dx
when x  .8, y  .6
dt
then s = 1
Figure 2.45:
The conical tank in Example 5.
Related
Rates
Water runs into a conical tank at the rate of 9 ft3/min. The
tank stands point down and has a height of 10 ft and a base of
radius 5 ft. How fast is the water level rising when the water
is 6 ft. deep?
Given:
dV
 9 ft 3 / min
dt
H  10 ft , R  5 ft
Find:
dy
when y  6 ft.
dt
Water
a conical
rate of 95.ft3/min. The
Figureruns
2.45:into
The
conicaltank
tankatinthe
Example
tank stands point down and has a height of 10 ft and a base of
radius 5 ft. How fast is the water level rising when the water
is 6 ft. deep?
1 2
V  x y
3
1 2
2 3
V   x (2 x)   x
3
3
dV
dx
 2 x 2
dt
dt
dx
9  2 (3) 2
dt
dx 1

dt 2
x 5

y 10
dy
dx
1 1
 2*  2*
  .32 ft / min
dt
dt
2 
y  2x
Given:
dV
 9 ft 3 / min
dt
H  10 ft , R  5 ft
Find:
dy
when y  6 ft.
dt
x=3
3. .10 The more we magnify the graph of a function near a
point where the function is differentiable, the flatter the graph
becomes and the more it resembles its tangent.
Differentiability
Differentiability and Linearization
Approximating the change in the function f by
the change in the tangent line of f.
Linearization
Write the equation of the straight line
approximation y  x  cos x at (0,1)
y  1  sin x  1  0  1 at (0,1)
y  y1  m( x  x1 )
y  f (a)  f (a)( x  a )
y  1  1( x  0)
y  1 x
Point-slope formula
y=f(x)