1. §2.6 Implicit Differentiation
Chabot Mathematics
§2.6 Implicit Differentiation
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. 2.5 Review § Any QUESTIONS About Any QUESTIONS About HomeWork
§2.5 → MarginalAnalysis and Increments
Any QUESTIONS About HomeWork
§2.5 → HW-11

3. §2.6 Learning Goals
Use implicit differentiation to find slopes and Rates of Change
Examine applied problems involving related rates of change

4. ReCall the Chain Rule
If f(u) is a differentiable fcn of u, and u(x) is a differentiable fcn of x, then
That is, the derivative of the composite function is the derivative of the “outside” function times the derivative of the “inside” function.

5. 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?

6. Example  Implicit Differentiation
If y = y(x) Then Find dy/dx from:
y(x) can NOT be algebraically isolated in this Expression (darn!)
Work-Around the Lack of Isolation using IMPLICIT Differentiation
Do on Doc Camera

7. 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
Consider the equation 2x2 + y2 = 8, the graph of which is an ellipse in the xy-plane

8. Comparison: Implicit vs Direct
For the Expression 2x2 + y2 = 8
Compute dy/dx by isolating y in the equation and then differentiating
Compute dy/dx by differentiating each term in the equation with respect to x and then solving for the derivative of y.
Compare the Two Results

9. 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([ ]); % 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',[ ], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', ' ','FontSize',7)
hold on
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
MATLAB Code

10. Comparison: Implicit vs Direct
SOLUTION (a)
First Isolate y:
Now differentiate with respect to x using the Chain Rule
Thus Ans

11. 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”.

12. Comparison: Implicit vs Direct
Using the implicit differentiation strategy, first differentiate each term in the equation:
Then
Now solve for the dy/dx term
Thus Ans

13. Comparison: Implicit vs Direct
SOLUTION - Comparison
Although the answers to parts (a) and (b) may look different, they should (and DO) agree:
Part (a)
Part (b)

14. 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?

15. Example  Crystal Growth
The most challenging part of this question is correctly identifying variables whose value(s) we need and variables whose value(s) we already know.
First, carefully examine the question
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?

16. Example  Crystal Growth
SOLUTION
Because the crystal is a cube, we know that V = s3
Now differentiate the volume equation with respect to time, using the chain rule (because volume and side length both depend on t):

17. Example  Crystal Growth
Need to Evaluate dV/dt when s = 3
Recall that the side length is growing at an instantaneous rate of 0.3 mm per hour; that is:
Then since
From Previous slide:

18. Example  Crystal Growth
State: When the sides are 3mm long, the sodium Choloride crystal is growing at a rate of 8.1 cubic millimeters per hour.

19. Related Rates 𝑑𝑧 𝑑𝑡 = 𝑑𝑧 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡 𝑑𝑧 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡 = 𝑑𝑧 𝑑𝑡
Related rates consist of using the CHAIN RULE BackWards
The Chain Rule:
Related Rates
𝑑𝑧 𝑑𝑡 = 𝑑𝑧 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡
𝑑𝑧 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡 = 𝑑𝑧 𝑑𝑡

20. 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
But r, S, and V are related by Geometry

21. Related Rates
Knowing u(t), v(t), and w(t) should allow calculation of quantities such as:
Consider a quick Example.
A 52 inch radius sphere expands at a rate of 3.7 inch/minute. Find dS/dV for these conditions
Recognize

22. Related Rates Employ the Chain Rule as Note that
Thus now have numbers for both dr/dt and dt/dr

23. Related Rates Find dS/dr by Direct Differentiation
Calc dr/dV by Implicit Differentiation

24. Related Rates
Solving for dr/dV
When r0 = 52 in, and dr/dt= 3.7 in/min

25. Related Rates
Recall So =1 =1

26. Example  Revenue vs. Time
The demand model for a product as a price fcn →
Where
D ≡ Demand in k-Units (kU)
x ≡ Product Price in $k/Unit
The price of the item decreases over time as
Where: t ≡ Elapsed Time after Product Release in Years (yr)

27. Example  Revenue vs. Time
Given D(x) & x(t) find the rate at which Revenue changes 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

28. Example  Revenue vs. Time
ReCall Revenue Definition
[Revenue] = [Demand]·[Quantity]
Mathematically in this case
The Above states R as fcn of x, but we need dR/dt
Can Use Related-Rates to eliminate x in Favor of t

29. Example  Revenue vs. Time
Use the ChainRule to determine dR/dt: Or
Now Use Product Rule on SqRt Term

30. Example  Revenue vs. Time
Continuing the ReDuction
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)

31. Example  Revenue vs. Time
Recall:
Then:
Can Now Calc dR/dt at the 6mon mark
State: After 6 months, revenue is increasing at a rate of about $1.16M per year (k-Units/year times $k/Unit)
k·k=M

32. WhiteBoard Work Problems From §2.6
P44 → Manufacturing Input-Compensation
P58 → Adiabatic Chemistry
P60 → Melting Ice

33. I Understand Implicitly
All Done for Today
I Understand Implicitly

35. 𝑑𝑦 𝑑𝑥

37. P2.6-38

49. P2.6-44

50. for

51. P2.6-58