Chabot Mathematics
§2.6 Implicit
Differentiation
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Review §
2.5
 Any QUESTIONS About
• §2.5 → MarginalAnalysis and
Increments
 Any QUESTIONS
About HomeWork
• §2.5 → HW-11
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
§2.6 Learning Goals
 Use implicit
differentiation to
find slopes and
Rates of Change
 Examine applied
problems involving
related rates of
change
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
ReCall the Chain Rule
 If f(u) is a differentiable fcn of u, and
u(x) is a differentiable fcn of x, then
df du df
f '  x   f ' u  x   f ' u   u '  x  


du dx dx
 That is, the derivative of the composite
function is the derivative of the “outside”
function times the derivative of the
“inside” function.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Implicit Differentiation
 Implicit differentiation is the process of
computing the derivative of the terms on
BOTH sides of an equation.
 This method is usually employed to find
the derivative of a dependent variable
when it is difficult or impossible to
isolate the dependent variable itself.
• This Typically Occurs for MULTIvariable
expressions; e.g., x·y(x) + [y(x)]1/2 = x3 − 23
– Then What is dy/dx?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Comparison: Implicit vs Direct
 In the x·y(x) + [y(x)]1/2 = x3 − 23 Problem
y(x) could NOT be isolated
algebraically; we HAD to use Impilicit
Differentiation to find dy/dx
• Sometimes, however, there is a choice
y
 Consider the equation
2x2 + y2 = 8, the graph
of which is an ellipse
in the xy-plane
MTH15 • 2x2 + y2 = 8 Ellipse
3
2
1
0
-1
-2
-3
-2.5
Chabot College Mathematics
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-2
-1.5
-1
-0.5
0
0.5
1
1.5
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
2
2.5
Comparison: Implicit vs Direct
 For the Expression 2x2 + y2 = 8
a) Compute dy/dx by isolating y in the
equation and then differentiating
b) Compute dy/dx by differentiating each
term in the equation with respect to x and
then solving for the derivative of y.
MTH15 • 2x2 + y2 = 8 Ellipse
3
 Compare the
Two Results
2
y
1
0
-1
-2
-3
-2.5
Chabot College Mathematics
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-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
2.5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
set(gca,'XTick',[xmin:.5:xmax]); set(gca,'YTick',[ymin:1:ymax])
plot([0,0],[ymin,ymax], 'k', [xmin, xmax], [0,0], 'k', 'LineWidth',
2)
hold off
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 08Jul13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
%
% The Limits
xmin = -2.5; xmax = 2.5;
ymin =-3; ymax =3;
% The FUNCTION
x = linspace(xmin+0.5,xmax-0.5,500); y1 = sqrt(8-2*x.^2); y2 = sqrt(8-2*x.^2);
%
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];
%
% the 6x6 Plot
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
axes; set(gca,'FontSize',12);
plot(x,y1,'b', x,y2, 'b', 'LineWidth', 4),axis([xmin xmax ymin
ymax]),...
grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y'),...
title(['\fontsize{16}MTH15 • 2x^2 + y^2 = 8 Ellipse',]),...
annotation('textbox',[.15 .05 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', ' ','FontSize',7)
hold on
Example  Implicit Differentiation
 If y = y(x) Then Find dy/dx from:
x  yx  x  yx  23
3
 y(x) can NOT be
algebraically isolated
in this Expression (darn!)
• Work-Around the Lack of
Isolation using IMPLICIT
Differentiation
Chabot College Mathematics
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Do on
y x 

x
White


x

y

d x
Board
dx

3
2

3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Comparison: Implicit vs Direct
 SOLUTION (a)
 First Isolate y:
2x 2 + y 2 = 8
y2 = 8- 2x 2
y = ± 8 - 2x 2
 Now differentiate
with respect to x:
Chabot College Mathematics
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(
)
dy d
=
± 8 - 2x 2
dx dx
1
2 1 / 2
  8  2 x    4 x 
2
2x

2 1/ 2
8  2 x 
 Thus Ans
dy
2x

2
dx
8  2x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Comparison: Implicit vs Direct
 SOLUTION (b)
 This last step is where the challenge (and
value) of implicit differentiation arises.
Each term is differentiated with x as its
input, so we carefully consider that y is
itself an expression that depends on x
• Thus, when we compute d(y2)/dx think of
chain rule and how “the square of y” is really
“the square of something with x’s in it”.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Comparison: Implicit vs Direct
 Using the implicit
differentiation
strategy, first
differentiate each
term in the equation:
d
d
2
2
2x + y = (8)
dx
dx
d
d 2
d
2
2x +
y = (8)
dx
dx
dx
d 2 dy
4x 
y
0
dy
dx
(
)
( )
( )
 
Chabot College Mathematics
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 Then 4x + 2y × dy = 0
dx
 Now solve for the
dy/dx term
dy
2y × = -4x
dx
dy -4x
=
.
dx 2y
 Thus Ans
dy  4 x

dx
2y
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Comparison: Implicit vs Direct
 SOLUTION - Comparison
 Although the answers to parts (a) and
(b) may look different, they should (and
DO) agree:
• Part (a)
-2x
dy -4x
=
• Part (b)
=
1/2
2
dx 2y ± (8 - 2x )
Chabot College Mathematics
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ü
ï
ïï
ý identical
ï
ï
ïþ
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Crystal Growth
 A sodium chloride crystal (c.f. ENGR45)
grows in the shape of a cube, with its
side lengths increasing by about 0.3
mm per hour.
 At what rate does the volume of the
rock salt crystal grow with respect
to time when the cube is 3 mm on a
side?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Crystal Growth
 The most challenging part of this
question is correctly identifying
variables whose value we need and
variables whose value we know.
 First, carefully examine the question
derivative
V
At what rate does the volume of the rock
salt crystal grow with respect to time when
d
the cube is 3 mm on a side?
Chabot College Mathematics
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s=3
dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Crystal Growth
 SOLUTION
 Because the crystal is a cube, we know
that V = s3
 Now differentiate the volume equation
with respect to time, d V   d s 3 
dt
using the chain rule dt
d
d 3 ds
(because volume
V   s 
dt
ds
dt
and side length
dV
both depend on t):
2 d
st 
 3s
Chabot College Mathematics
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dt
dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Crystal Growth
 Need to Evaluate dV/dt when s = 3
 Recall that the side length is growing at
ds
an instantaneous rate of
 0.3
0.3 mm per hour; that is: dt
 Then since dV  3s 2 d st   3s 2  ds
dt
dV
dt
Chabot College Mathematics
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s 3
ds dt  0.3
dt
dt
3
mm
2
 33  0.3  8.1
Hr
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Crystal Growth
 State: When the sides are dV
dt
3mm long, the sodium
s 3
ds dt  0.3
mm 3
 8.1
Hr
Choloride crystal is growing at a rate
of 8.1 cubic millimeters per hour.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Related Rates
 In many situations two, or more, rates
(derivatives), are related in Some Way.
 Example Consider a Sphere Expanding
in TIME with radius, r(t), Surface area,
S(t), and Volume, V(t), then
dr
dS
dV
 u t 
 vt 
 wt 
dt
dt
dt
 But r, S, and V are related by Geometry
S  4r
Chabot College Mathematics
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2
4 3
V  r
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Related Rates
 Knowing u(t), v(t), and w(t)
dV
should allow calculation
dr
of quantities such as:
 Consider a quick Example.
dS
dr
• A 52 inch radius sphere expands at a rate
of 3.7 inch/minute. Find dS/dV for these
conditions
• Recognize r0  52 in
Chabot College Mathematics
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dr
dt
r0
dV
dS
in
 3.7
min
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Related Rates
 Employ the Chain Rule as
dS  dS   dt   dS dr   dt dr 
        
dV  dt   dV   dr dt   dr dV 
 Note that
dr
in
 3.7
dt
min
dt
1
1
min



 0.2703
dr dt 3.7 in
in
dr 1 min
 Thus now have numbers for both
dr/dt and dt/dr
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Related Rates
 Find dS/dr by Direct Differentiation
2
d
d
in
dS
2
S  4r  4 2r  8r
 8r in 
dr
dr
in
dr
 Calc dr/dV by Implicit Differentiation
4 3
d 
4 3 
V  r V  
V  r V 

3
dV 
3

d
d 4 3
4 d 3
V
r V   1  
r
dV
dV 3
3 dV
4 d 3
4 d 3 dr
4
2 dr
1 
r  1 
r
 1   3r
3 dV
3 dr dV
3
dV
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Related Rates
 Solving
1
dr
2 dr
1  4r


2
for dr/dV
dV
4r
dV
 When r0 = 52 in, and dr/dt= 3.7 in/min
dS
dr
dr
dV
r0
r0
1
1
3 1


 5.813 10
2
2
2
in
4 r0 
4 3.7 
Chabot College Mathematics
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 8r0  8 3.7   92.99 in
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Related Rates
 Recall
dS  dS dr   dt dr  dS dr dt dr
    




dV  dr dt   dr dV  dr dt dr dV
 So
dS
in 3.7 in
1 min
3 1
 92.99 

 5.813 10
dV r
1 1 min 3.7 in
in 2
0
dS
dV
in
1
 92.99  5.813 10
 0.5405
3
in
in
3
r0
Chabot College Mathematics
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2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Revenue vs. Time
 The demand model for a Dx   4  2 x  1
product as a function →
• Where
– D ≡ Demand in k-Units (kU)
– x ≡ Product Price in $k/Unit
 The price of the item
decreases over time as
xt   6  t 2
• Where: t ≡ Time after Product Release in
Years (yr)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Revenue vs. Time
 Given D(x) & x(t) at what rate is
Revenue changing with respect to time
six months after the item’s release?
 SOLUTION
 Formalizing the goal with mathematics,
we want to know the rate, dR/dt , six
months after release.
• Because time is measured in years,
set t = 0.5 years
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Revenue vs. Time
 ReCall Revenue Definition
[Revenue] = [Demand]·[Quantity]
 Mathematically in this case
(
)
Rx  x  Dx = x × 4 - 2x +1  4 x  x 2 x  1
 The Above states R as fcn of x, but we
need dR/dt
• Can Use Related-Rates to eliminate
x in Favor of t
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Revenue vs. Time
 Use the ChainRule to determine dR/dt:


dR  dR   dx   d
d



     
4x  x 2x 1    6  t 2 
dt  dx   dt   dx
  dt


 Or dR  d



 4 
x 2 x  1    2t 
dt

dx

 Now Use Product Rule on SqRt Term

d
x 2x 1
dx
Chabot College Mathematics
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
1
1 / 2
 x   2 x  1  2  1  2 x  1
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx

Example  Revenue vs. Time
 Continuing the ReDuction
ù
dR é
1
-1/2
= ê4 - x × ( 2x +1) × 2 -1× 2x +1ú × (-2t )
û
dt ë
2
-1/2
é
= ë4 - x ( 2x +1) - 2x +1ùû × (-2t )
 We need to evaluate the revenue
derivative at t = 0.5 yrs, but there’s a
catch: We know the value of t, but the
value of x is not explicitly known.
• Use the Price Fcn to calculate x0 = x(0.5yr)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
Example  Revenue vs. Time
 Recall: xt   6  t 2
2
 Then: x0  xt0   x5  6  0.5  5.75
 Can Now Calc dR/dt at the 6mon mark


dR
1/ 2
 4  5.7525.75  1  25.75  1   2  0.5
dt t 0.5, x 5.75
 1.162 $M yr
• State: After 6 months, revenue is
increasing at a rate of about $1.162M per
year (k-Units/year times $k/Unit)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
WhiteBoard Work
 Problems From §2.6
• P44 → Manufacturing Input-Compensation
• P58 → Adiabatic Chemistry
• P60 → Melting Ice
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
All Done for Today
I
Understand
Implicitly
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
P2.6-44
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
P2.6-58
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx