1
Example Find
dy
if x3 + 4xy2 − 7 = y3.
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7)
dx
=
d 3
y
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7)
dx
3x2 +
=
d 3
y
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7)
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x −
dx
dx
=
d 3
y
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
d 3
y
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
d 3
y
dx
3y2 ·
dy
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
3x2 + 4x · 2y ·
dy
+
dx
d 3
y
dx
3y2 ·
dy
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
!
dy
3x2 + 4x · 2y ·
+ y2 · 4 =
dx
d 3
y
dx
3y2 ·
dy
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
!
dy
3x2 + 4x · 2y ·
+ y2 · 4 =
dx
d 3
y
dx
3y2 ·
3y2
dy
dx
dy
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
!
dy
3x2 + 4x · 2y ·
+ y2 · 4 =
dx
3x2 + 8xy
dy
+ 4y2
dx
=
d 3
y
dx
3y2 ·
dy
dx
3y2
dy
dx
3y2
dy
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
!
dy
3x2 + 4x · 2y ·
+ y2 · 4 =
dx
3x2 + 8xy
dy
+ 4y2
dx
2
2
3x + 4y
d 3
y
dx
3y2 ·
dy
dx
3y2
dy
dx
=
3y2
dy
dx
=
dy
(3y − 8xy)
dx
2
dy
solve for
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
!
dy
3x2 + 4x · 2y ·
+ y2 · 4 =
dx
3x2 + 8xy
dy
+ 4y2
dx
2
2
3x + 4y
Compare
3a + 8bz + 4c = 3cz
d 3
y
dx
3y2 ·
dy
dx
3y2
dy
dx
=
3y2
dy
dx
=
dy
(3y − 8xy)
dx
2
dy
solve for
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
!
dy
3x2 + 4x · 2y ·
+ y2 · 4 =
dx
3x2 + 8xy
dy
+ 4y2
dx
2
2
3x + 4y
Compare
3a + 8bz + 4c = 3cz
d 3
y
dx
3y2 ·
dy
dx
3y2
dy
dx
=
3y2
dy
dx
=
dy
(3y − 8xy)
dx
2
3a + 4c = 3cz − 8bz
dy
solve for
dx
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
!
dy
3x2 + 4x · 2y ·
+ y2 · 4 =
dx
3x2 + 8xy
dy
+ 4y2
dx
2
2
3x + 4y
Compare
3a + 8bz + 4c = 3cz
d 3
y
dx
3y2 ·
dy
dx
3y2
dy
dx
=
3y2
dy
dx
=
dy
(3y − 8xy)
dx
2
3a + 4c = 3cz − 8bz
dy
solve for
dx
3a + 4c = (3c − 8b)z
1
Example Find
Solution
dy
if x3 + 4xy2 − 7 = y3.
dx
Differentiate both sides of the given equation w.r.t. x
d 3
(x + 4x · y2 − 7) =
dx
!
d 2
d
3x2 + 4x ·
y + y2 ·
4x − 0 =
dx
dx
!
dy
3x2 + 4x · 2y ·
+ y2 · 4 =
dx
3x2 + 8xy
dy
+ 4y2
dx
2
2
3x + 4y
dy
dx
Compare
3a + 8bz + 4c = 3cz
d 3
y
dx
3y2 ·
dy
dx
3y2
dy
dx
=
3y2
dy
dx
=
dy
(3y − 8xy)
dx
=
3x2 + 4y2
3y2 − 8xy
2
3a + 4c = 3cz − 8bz
dy
solve for
dx
3a + 4c = (3c − 8b)z
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
d
(x · ey − 1)
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
d
(x · ey − 1)
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x −
dx
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x − 0
dx
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
y·
1
+
x
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x − 0
dx
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
y·
1
dy
+ ln x ·
x
dx
=
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x − 0
dx
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
y·
1
dy
+ ln x ·
x
dx
=
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x − 0
dx
dx
x · ey ·
dy
+
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
y·
1
dy
+ ln x ·
x
dx
=
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x − 0
dx
dx
x · ey ·
dy
+ ey · 1
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
y·
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x − 0
dx
dx
1
dy
+ ln x ·
x
dx
=
x · ey ·
dy
dx
=
x2ey
y + x ln x
dy
+ ey · 1
dx
dy
+ x ey
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
y·
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x − 0
dx
dx
1
dy
+ ln x ·
x
dx
=
x · ey ·
dy
dx
=
x2ey
=
y
y + x ln x
dy
(x ln x − x e )
dx
2 y
dy
+ ey · 1
dx
dy
+ x ey
dx
xe − y
dy
solve for
dx
2
Example Find
Solution
dy
if y ln x = xey − 1.
dx
Differentiate both sides w.r.t. x
d
(y · ln x) =
dx
y·
d
d
ln x + ln x · y =
dx
dx
y·
d
(x · ey − 1)
dx
x·
d y
d
e + ey · x − 0
dx
dx
1
dy
+ ln x ·
x
dx
=
x · ey ·
dy
dx
=
x2ey
=
y
xe − y
=
x ey − y
x ln x − x2ey
y + x ln x
dy
(x ln x − x e )
dx
2 y
dy
dx
dy
+ ey · 1
dx
dy
+ x ey
dx
dy
solve for
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
2
x · sin y + cos (y ) =
dx
d
1
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
2
x · sin y + cos (y ) =
dx
!
d
d
x · sin y + sin y · x +
dx
dx
d
1
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
2
x · sin y + cos (y ) =
dx
!
d 2
d
d
2
x · sin y + sin y · x + (− sin y ) · y
dx
dx
dx
d
1
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
d
2
x · sin y + cos (y ) =
1
dx
dx
!
d 2
d
d
2
x · sin y + sin y · x + (− sin y ) · y = 0
dx
dx
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
d
2
x · sin y + cos (y ) =
1
dx
dx
!
d 2
d
d
2
x · sin y + sin y · x + (− sin y ) · y = 0
dx
dx
dx
x · cos y ·
dy
+
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
d
2
x · sin y + cos (y ) =
1
dx
dx
!
d 2
d
d
2
x · sin y + sin y · x + (− sin y ) · y = 0
dx
dx
dx
!
dy
x · cos y ·
+ sin y · 1
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
d
2
x · sin y + cos (y ) =
1
dx
dx
!
d 2
d
d
2
x · sin y + sin y · x + (− sin y ) · y = 0
dx
dx
dx
!
dy
dy
x · cos y ·
+ sin y · 1 − sin y2 · 2y ·
dx
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
d
2
x · sin y + cos (y ) =
1
dx
dx
!
d 2
d
d
2
x · sin y + sin y · x + (− sin y ) · y = 0
dx
dx
dx
!
dy
dy
x · cos y ·
+ sin y · 1 − sin y2 · 2y ·
= 0
dx
dx
3
Example Find the slope of the curve with equation
x sin y + cos y2 = 1
at the point (1, 0).
Solution
• Differentiate the given equation w.r.t x
d
d
2
x · sin y + cos (y ) =
1
dx
dx
!
d 2
d
d
2
x · sin y + sin y · x + (− sin y ) · y = 0
dx
dx
dx
!
dy
dy
x · cos y ·
+ sin y · 1 − sin y2 · 2y ·
= 0
dx
dx
x cos y
dy
dy
+ sin y − 2y sin y2
dx
dx
= 0
(∗)
4
dy
• Solve for
dx
4
dy
• Solve for
dx
dy
sin y = (2y sin y − x cos y)
dx
2
4
dy
• Solve for
dx
dy
sin y = (2y sin y − x cos y)
dx
2
dy
dx
=
sin y
2y sin y2 − x cos y
4
dy
• Solve for
dx
dy
sin y = (2y sin y − x cos y)
dx
2
dy
dx
=
sin y
2y sin y2 − x cos y
• The slope of the curve at (1, 0) is
dy dx (1,0)
4
dy
• Solve for
dx
dy
sin y = (2y sin y − x cos y)
dx
2
dy
dx
=
sin y
2y sin y2 − x cos y
• The slope of the curve at (1, 0) is
dy dx (1,0)
=
sin 0
0 − 1 · cos 0
4
dy
• Solve for
dx
dy
sin y = (2y sin y − x cos y)
dx
2
dy
dx
=
sin y
2y sin y2 − x cos y
• The slope of the curve at (1, 0) is
dy dx (1,0)
=
sin 0
0 − 1 · cos 0
= 0
4
dy
• Solve for
dx
dy
sin y = (2y sin y − x cos y)
dx
2
dy
dx
=
sin y
2y sin y2 − x cos y
• The slope of the curve at (1, 0) is
dy dx (1,0)
=
sin 0
0 − 1 · cos 0
= 0
Remark Alternatively, put (x, y) = (1, 0) into (∗)
4
dy
• Solve for
dx
dy
sin y = (2y sin y − x cos y)
dx
2
dy
dx
=
sin y
2y sin y2 − x cos y
• The slope of the curve at (1, 0) is
dy dx (1,0)
=
sin 0
0 − 1 · cos 0
= 0
Remark Alternatively, put (x, y) = (1, 0) into (∗)
dy
dy
1 · cos 0 · + sin 0 − 2 · 0 · sin 02 · = 0
dx (1,0)
dx (1,0)
4
dy
• Solve for
dx
dy
sin y = (2y sin y − x cos y)
dx
2
dy
dx
=
sin y
2y sin y2 − x cos y
• The slope of the curve at (1, 0) is
dy dx (1,0)
=
sin 0
0 − 1 · cos 0
= 0
Remark Alternatively, put (x, y) = (1, 0) into (∗)
dy
dy
1 · cos 0 · + sin 0 − 2 · 0 · sin 02 · = 0
dx (1,0)
dx (1,0)
dy + 0 − 0 = 0
dx (1,0)
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
Let r(t) = radius of circle at time t
A(t) = area
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
Let r(t) = radius of circle at time t
A(t) = area
A = πr2
(1)
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
Let r(t) = radius of circle at time t
A(t) = area
A = πr2
dr
Given
= 3.
dt
(1)
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
Let r(t) = radius of circle at time t
A(t) = area
A = πr2
dr
Given
= 3.
dt
Find
dA
when r = 5, that is,
dt
(1)
5
Example The radius of a circle is increasing at the rate of 3 cm per second.
Find the rate of change of the area inside the circle when the radius is 5 cm.
Solution
Let r(t) = radius of circle at time t
A(t) = area
A = πr2
(1)
dr
Given
= 3.
dt
Find
dA
when r = 5, that is,
dt
dA where t0 is the time at which r(t0) = 5
find
dt t=t0
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
A =
dt
d 2
πr
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
A =
dt
dA
=
dt
d 2
πr
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
d 2
A =
πr
dt
dt
dr
dA
= π · 2r
dt
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
d 2
A =
πr
dt
dt
dr
dA
= π · 2r
dt
dt
d
d
2
f (t) = 2 f (t) f (t)
dt
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
d 2
A =
πr
dt
dt
dr
dA
= π · 2r
dt
dt
= π · 2r · 3
d
d
2
f (t) = 2 f (t) f (t)
dt
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
d 2
A =
πr
dt
dt
dr
dA
= π · 2r
dt
dt
= π · 2r · 3 = 6πr
d
d
2
f (t) = 2 f (t) f (t)
dt
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
d 2
A =
πr
dt
dt
dr
dA
= π · 2r
dt
dt
= π · 2r · 3 = 6πr
At time t0
dA dt t=t0
= 6π · r(t0)
d
d
2
f (t) = 2 f (t) f (t)
dt
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
d 2
A =
πr
dt
dt
dr
dA
= π · 2r
dt
dt
= π · 2r · 3 = 6πr
At time t0
dA dt t=t0
= 6π · r(t0)
= 6π · 5
d
d
2
f (t) = 2 f (t) f (t)
dt
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
d 2
A =
πr
dt
dt
dr
dA
= π · 2r
dt
dt
= π · 2r · 3 = 6πr
At time t0
dA dt t=t0
= 6π · r(t0)
= 6π · 5
= 30π
d
d
2
f (t) = 2 f (t) f (t)
dt
dt
6
Solution (cont.)
Differentiate (1) with respect to time t:
d
d 2
A =
πr
dt
dt
dr
dA
= π · 2r
dt
dt
d
d
2
f (t) = 2 f (t) f (t)
dt
dt
= π · 2r · 3 = 6πr
At time t0
dA dt t=t0
= 6π · r(t0)
= 6π · 5
= 30π
The area is increasing at the rate of 30π cm2 per second.
7
Related Rates
7
Related Rates
• Obtain/given an equation relating two quantities x and y, both of them are
functions of time t.
7
Related Rates
• Obtain/given an equation relating two quantities x and y, both of them are
functions of time t.
• Differentiate the equation with respect to t.
7
Related Rates
• Obtain/given an equation relating two quantities x and y, both of them are
functions of time t.
• Differentiate the equation with respect to t.
• Get an equation involving x, y,
dx
dy
and .
dt
dt
7
Related Rates
• Obtain/given an equation relating two quantities x and y, both of them are
functions of time t.
• Differentiate the equation with respect to t.
• Get an equation involving x, y,
• Plugin given data.
dx
dy
and .
dt
dt
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
Solution
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
1
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
d
(4x2 + y2) =
dt
d
8
dt
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
1
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
d
(4x2 + y2) =
dt
dx
4 · 2x ·
+
dt
d
8
dt
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
d
d
2
2
(4x + y ) =
8
dt
dt
dx
dy
4 · 2x ·
+ 2y ·
= 0
dt
dt
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
d
d
2
2
(4x + y ) =
8
dt
dt
dx
dy
4 · 2x ·
+ 2y ·
= 0
dt
dt
dx
dy
4x
+y
= 0
dt
dt
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
d
d
2
2
(4x + y ) =
8
dt
dt
dx
dy
4 · 2x ·
+ 2y ·
= 0
dt
dt
dx
dy
4x
+y
= 0
dt
dt
At that moment
x = 1, y = 2 and
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
At that moment
d
d
2
2
(4x + y ) =
8
dt
dt
dx
dy
4 · 2x ·
+ 2y ·
= 0
dt
dt
dx
dy
4x
+y
= 0
dt
dt
dx
= 3.
x = 1, y = 2 and
dt
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
At that moment
d
d
2
2
(4x + y ) =
8
dt
dt
dx
dy
4 · 2x ·
+ 2y ·
= 0
dt
dt
dx
dy
4x
+y
= 0
dt
dt
dx
= 3.
x = 1, y = 2 and
dt
4·1·3+
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
1
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
At that moment
d
(4x2 + y2) =
dt
dx
dy
4 · 2x ·
+ 2y ·
=
dt
dt
dx
dy
4x
+y
=
dt
dt
dx
= 3.
x = 1, y = 2 and
dt
dy
= 0
4·1·3+2
dt
d
8
dt
0
0
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
At that moment
d
d
2
2
(4x + y ) =
8
dt
dt
dx
dy
4 · 2x ·
+ 2y ·
= 0
dt
dt
dx
dy
4x
+y
= 0
dt
dt
dx
= 3.
x = 1, y = 2 and
dt
dy
= 0
4·1·3+2
dt
dy
= −6
dt
1
8
Example
A point is moving along the graph of
4x2 + y2 = 8. When the point is at (1, 2), its x-coordinate
2
1
is increasing at the rate of 3 units per second. How fast
is the y-coordinate changing at that moment?
Solution Differentiate equation of ellipse w.r.t. time t
d
d
2
2
(4x + y ) =
8
dt
dt
dx
dy
4 · 2x ·
+ 2y ·
= 0
dt
dt
dx
dy
4x
+y
= 0
dt
dt
dx
= 3.
At that moment x = 1, y = 2 and
dt
dy
= 0
4·1·3+2
dt
dy
= −6
dt
The y-coordinate is decreasing at the rate of 6 units per second.
1
9
More graph sketching
Example Let f (x) = x ln x.
(1) Find and classify the critical point(s) of f .
(2) Find the interval(s) on which f is increasing/decreasing, convex/concave.
(3) Sketch the graph of f .
Solution
9
More graph sketching
Example Let f (x) = x ln x.
(1) Find and classify the critical point(s) of f .
(2) Find the interval(s) on which f is increasing/decreasing, convex/concave.
(3) Sketch the graph of f .
Solution
• f (x) is defined for x > 0 only.
domain = (0, ∞)
9
More graph sketching
Example Let f (x) = x ln x.
(1) Find and classify the critical point(s) of f .
(2) Find the interval(s) on which f is increasing/decreasing, convex/concave.
(3) Sketch the graph of f .
Solution
• f (x) is defined for x > 0 only.
•
f 0(x) =
x·
d
d
ln x + ln x · x
dx
dx
domain = (0, ∞)
9
More graph sketching
Example Let f (x) = x ln x.
(1) Find and classify the critical point(s) of f .
(2) Find the interval(s) on which f is increasing/decreasing, convex/concave.
(3) Sketch the graph of f .
Solution
• f (x) is defined for x > 0 only.
•
f 0(x) =
=
x·
d
d
ln x + ln x · x
dx
dx
1
x · + ln x · 1
x
domain = (0, ∞)
9
More graph sketching
Example Let f (x) = x ln x.
(1) Find and classify the critical point(s) of f .
(2) Find the interval(s) on which f is increasing/decreasing, convex/concave.
(3) Sketch the graph of f .
Solution
• f (x) is defined for x > 0 only.
•
f 0(x) =
=
x·
d
d
ln x + ln x · x
dx
dx
1
x · + ln x · 1
x
= 1 + ln x
domain = (0, ∞)
9
More graph sketching
Example Let f (x) = x ln x.
(1) Find and classify the critical point(s) of f .
(2) Find the interval(s) on which f is increasing/decreasing, convex/concave.
(3) Sketch the graph of f .
Solution
• f (x) is defined for x > 0 only.
•
f 0(x) =
=
x·
d
d
ln x + ln x · x
dx
dx
1
x · + ln x · 1
x
= 1 + ln x
• Solve f 0(x) = 0
ln x = −1
domain = (0, ∞)
9
More graph sketching
Example Let f (x) = x ln x.
(1) Find and classify the critical point(s) of f .
(2) Find the interval(s) on which f is increasing/decreasing, convex/concave.
(3) Sketch the graph of f .
Solution
• f (x) is defined for x > 0 only.
•
f 0(x) =
=
x·
domain = (0, ∞)
d
d
ln x + ln x · x
dx
dx
1
x · + ln x · 1
x
= 1 + ln x
• Solve f 0(x) = 0
ln x = −1
x = e−1
critical point of f
10
0 < x < e−1
f 0(x) = 1 + ln x
f
x = e−1
0
x > e−1
10
f 0(x) = 1 + ln x
f
0 < x < e−1
x = e−1
−
0
x > e−1
10
f 0(x) = 1 + ln x
f
0 < x < e−1
x = e−1
x > e−1
−
0
+
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
%
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
%
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
−1
f has a local min at x = e .
%
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
−1
f has a local min at x = e .
•
00
f (x) =
d
(1 + ln x)
dx
%
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
−1
f has a local min at x = e .
•
00
f (x) =
=
d
(1 + ln x)
dx
1
x
%
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
−1
f has a local min at x = e .
•
00
f (x) =
=
d
(1 + ln x)
dx
1
x
f 00(x) = 0 has no solution
%
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
−1
f has a local min at x = e .
•
00
f (x) =
=
d
(1 + ln x)
dx
1
x
f 00(x) = 0 has no solution
• Since f 00(x) > 0 for all x > 0,
f is convex on (0, ∞).
%
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
−1
f has a local min at x = e .
•
00
f (x) =
=
d
(1 + ln x)
dx
1
x
f 00(x) = 0 has no solution
• Since f 00(x) > 0 for all x > 0,
f is convex on (0, ∞).
f (x)
0<x<1
x=1
x>1
−
0
+
%
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
−1
f has a local min at x = e .
•
00
f (x) =
=
%
d
(1 + ln x)
dx
1
x
f 00(x) = 0 has no solution
• Since f 00(x) > 0 for all x > 0,
f is convex on (0, ∞).
f (x)
1
0<x<1
x=1
x>1
−
0
+
1
E
1
2
10
0 < x < e−1
x = e−1
x > e−1
f 0(x) = 1 + ln x
−
0
+
f
&
−1
• f is decreasing on (0, e )
−1
increasing on (e , ∞)
−1
f has a local min at x = e .
•
00
f (x) =
=
%
d
(1 + ln x)
dx
1
x
f 00(x) = 0 has no solution
• Since f 00(x) > 0 for all x > 0,
f is convex on (0, ∞).
f (x)
1
0<x<1
x=1
x>1
−
0
+
Remark lim x ln x = 0
x→0+
1
E
1
2