Section 2.6Implicit Differentiation
V63.0121.021, Calculus I
New York University
October 12, 2010
Announcements
I Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2I Midterm next week. Covers §§1.1–2.5
. . . . . .
. . . . . .
Announcements
I Quiz 2 in recitation thisweek. Covers §§1.5, 1.6,2.1, 2.2
I Midterm next week.Covers §§1.1–2.5
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 2 / 34
. . . . . .
Objectives
I Use implicit differentationto find the derivative of afunction defined implicitly.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 3 / 34
. . . . . .
Outline
The big idea, by example
ExamplesBasic ExamplesVertical and Horizontal TangentsOrthogonal TrajectoriesChemistry
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 4 / 34
. . . . . .
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.
Solution (Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Why the −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
. . . . . .
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.
Solution (Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Why the −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
. . . . . .
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.
Solution (Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Why the −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
. . . . . .
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.
Solution (Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Why the −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
. . . . . .
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.
Solution (Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Why the −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
. . . . . .
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.
Solution (Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Why the −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
. . . . . .
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
. .x
.y
.
Solution (Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Why the −?)
I Differentiate:dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
. . . . . .
Motivating Example, another way
We know that x2 + y2 = 1 does not define y as a function of x, butsuppose it did.
I Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
I We could differentiate this equation to get
2x+ 2f(x) · f′(x) = 0
I We could then solve to get
f′(x) = − xf(x)
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
. . . . . .
Motivating Example, another way
We know that x2 + y2 = 1 does not define y as a function of x, butsuppose it did.
I Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
I We could differentiate this equation to get
2x+ 2f(x) · f′(x) = 0
I We could then solve to get
f′(x) = − xf(x)
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
. . . . . .
Motivating Example, another way
We know that x2 + y2 = 1 does not define y as a function of x, butsuppose it did.
I Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
I We could differentiate this equation to get
2x+ 2f(x) · f′(x) = 0
I We could then solve to get
f′(x) = − xf(x)
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
.looks like a function
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
.
.looks like a function
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
..does not look like afunction, but that’sOK—there are onlytwo points like this
.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
.looks like a function
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y2 = 1, thecurve resembles the graphof a function.
I So f(x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
. .x
.y
.
.looks like a function
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
. . . . . .
Motivating Example, again, with Leibniz notation
ProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0Remember y is assumed to be a function of x!
I Isolate:dydx
= −xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
. . . . . .
Motivating Example, again, with Leibniz notation
ProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0
Remember y is assumed to be a function of x!
I Isolate:dydx
= −xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
. . . . . .
Motivating Example, again, with Leibniz notation
ProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0Remember y is assumed to be a function of x!
I Isolate:dydx
= −xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
. . . . . .
Motivating Example, again, with Leibniz notation
ProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0Remember y is assumed to be a function of x!
I Isolate:dydx
= −xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
. . . . . .
Motivating Example, again, with Leibniz notation
ProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0Remember y is assumed to be a function of x!
I Isolate:dydx
= −xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
. . . . . .
Summary
If a relation is given between x and y which isn’t a function:
I “Most of the time”, i.e., “atmost places” y can beassumed to be a function of x
I we may differentiate therelation as is
I Solving fordydx
does give theslope of the tangent line to thecurve at a point on the curve.
. .x
.y
.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 9 / 34
. . . . . .
Outline
The big idea, by example
ExamplesBasic ExamplesVertical and Horizontal TangentsOrthogonal TrajectoriesChemistry
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 10 / 34
. . . . . .
Another Example
Example
Find y′ along the curve y3 + 4xy = x2 + 3.
SolutionImplicitly differentiating, we have
3y2y′ + 4(1 · y+ x · y′) = 2x
Solving for y′ gives
3y2y′ + 4xy′ = 2x− 4y
(3y2 + 4x)y′ = 2x− 4y
=⇒ y′ =2x− 4y3y2 + 4x
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
. . . . . .
Another Example
Example
Find y′ along the curve y3 + 4xy = x2 + 3.
SolutionImplicitly differentiating, we have
3y2y′ + 4(1 · y+ x · y′) = 2x
Solving for y′ gives
3y2y′ + 4xy′ = 2x− 4y
(3y2 + 4x)y′ = 2x− 4y
=⇒ y′ =2x− 4y3y2 + 4x
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
. . . . . .
Another Example
Example
Find y′ along the curve y3 + 4xy = x2 + 3.
SolutionImplicitly differentiating, we have
3y2y′ + 4(1 · y+ x · y′) = 2x
Solving for y′ gives
3y2y′ + 4xy′ = 2x− 4y
(3y2 + 4x)y′ = 2x− 4y
=⇒ y′ =2x− 4y3y2 + 4x
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
. . . . . .
Yet Another Example
Example
Find y′ if y5 + x2y3 = 1+ y sin(x2).
SolutionDifferentiating implicitly:
5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)
Collect all terms with y′ on one side and all terms without y′ on theother:
5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)
Now factor and divide:
y′ =2xy(cos x2 − y2)
5y4 + 3x2y2 − sin x2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 12 / 34
. . . . . .
Yet Another Example
Example
Find y′ if y5 + x2y3 = 1+ y sin(x2).
SolutionDifferentiating implicitly:
5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)
Collect all terms with y′ on one side and all terms without y′ on theother:
5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)
Now factor and divide:
y′ =2xy(cos x2 − y2)
5y4 + 3x2y2 − sin x2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 12 / 34
. . . . . .
Finding tangent lines with implicit differentitiation.
.
Example
Find the equation of the linetangent to the curve
y2 = x2(x+ 1) = x3 + x2
at the point (3,−6).
.
.
Solution
Differentiate: 2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thus the equation of the tangent line is y+ 6 = −114(x− 3).
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
. . . . . .
Finding tangent lines with implicit differentitiation.
.
Example
Find the equation of the linetangent to the curve
y2 = x2(x+ 1) = x3 + x2
at the point (3,−6).
.
.
Solution
Differentiate: 2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thus the equation of the tangent line is y+ 6 = −114(x− 3).
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
. . . . . .
Finding tangent lines with implicit differentitiation.
.
Example
Find the equation of the linetangent to the curve
y2 = x2(x+ 1) = x3 + x2
at the point (3,−6).
.
.
Solution
Differentiate: 2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thus the equation of the tangent line is y+ 6 = −114(x− 3).
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
. . . . . .
Recall: Line equation forms
I slope-intercept formy = mx+ b
where the slope is m and (0,b) is on the line.I point-slope form
y− y0 = m(x− x0)
where the slope is m and (x0, y0) is on the line.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 14 / 34
. . . . . .
Horizontal Tangent Lines
Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
SolutionWe have to solve these two equations:
..
y2 = x3 + x2
[(x, y) is on the curve].1.
3x2 + 2x2y
= 0
[tangent lineis horizontal]
.2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 15 / 34
. . . . . .
Horizontal Tangent Lines
Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
SolutionWe have to solve these two equations:
..
y2 = x3 + x2
[(x, y) is on the curve].1.
3x2 + 2x2y
= 0
[tangent lineis horizontal]
.2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 15 / 34
. . . . . .
Solution, continued
I Solving the second equation gives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.
I Substituting x = 0 into the first equation gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down that road.I Substituting x = −2/3 into the first equation gives
y2 =
(−23
)3+
(−23
)2=
427
=⇒ y = ± 23√3,
so there are two horizontal tangents.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
. . . . . .
Solution, continued
I Solving the second equation gives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.
I Substituting x = 0 into the first equation gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down that road.I Substituting x = −2/3 into the first equation gives
y2 =
(−23
)3+
(−23
)2=
427
=⇒ y = ± 23√3,
so there are two horizontal tangents.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
. . . . . .
Solution, continued
I Solving the second equation gives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.I Substituting x = 0 into the first equation gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down that road.
I Substituting x = −2/3 into the first equation gives
y2 =
(−23
)3+
(−23
)2=
427
=⇒ y = ± 23√3,
so there are two horizontal tangents.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
. . . . . .
Solution, continued
I Solving the second equation gives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.I Substituting x = 0 into the first equation gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down that road.I Substituting x = −2/3 into the first equation gives
y2 =
(−23
)3+
(−23
)2=
427
=⇒ y = ± 23√3,
so there are two horizontal tangents.V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
. . . . . .
Horizontal Tangents
..
.(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
)
.
.node
..(−1,0)
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 17 / 34
. . . . . .
Horizontal Tangents
..
.(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
) .
.node
..(−1,0)
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 17 / 34
. . . . . .
Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
I Tangent lines are vertical whendxdy
= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(notice this is the
reciprocal of dy/dx).I We must solve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0[tangent lineis vertical]
.2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
. . . . . .
Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
I Tangent lines are vertical whendxdy
= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(notice this is the
reciprocal of dy/dx).I We must solve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0[tangent lineis vertical]
.2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
. . . . . .
Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
I Tangent lines are vertical whendxdy
= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(notice this is the
reciprocal of dy/dx).
I We must solve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0[tangent lineis vertical]
.2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
. . . . . .
Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
I Tangent lines are vertical whendxdy
= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(notice this is the
reciprocal of dy/dx).I We must solve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0[tangent lineis vertical]
.2V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
. . . . . .
Solution, continued
I Solving the second equation gives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(as long as 3x2 + 2x ̸= 0).
I Substituting y = 0 into the first equation gives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.I x = 0 is not allowed by the first equation, but
dxdy
∣∣∣∣(−1,0)
= 0,
so here is a vertical tangent.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
. . . . . .
Solution, continued
I Solving the second equation gives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(as long as 3x2 + 2x ̸= 0).I Substituting y = 0 into the first equation gives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.
I x = 0 is not allowed by the first equation, but
dxdy
∣∣∣∣(−1,0)
= 0,
so here is a vertical tangent.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
. . . . . .
Solution, continued
I Solving the second equation gives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(as long as 3x2 + 2x ̸= 0).I Substituting y = 0 into the first equation gives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.I x = 0 is not allowed by the first equation, but
dxdy
∣∣∣∣(−1,0)
= 0,
so here is a vertical tangent.V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
. . . . . .
Tangents
..
.(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
) .
.node
..(−1,0)
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 20 / 34
. . . . . .
Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx
I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy
I The product is −1, so the tangent lines are perpendicularwherever they intersect.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 21 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Orthogonal Families of Curves
xy = cx2 − y2 = k . .x
.y
.xy=1
.xy=2
.xy=3
.xy=−1
.xy=−2
.xy=−3
.x2−
y2=1
.x2−
y2=2
.x2−
y2=3
.x2 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
. . . . . .
Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx
I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy
I The product is −1, so the tangent lines are perpendicularwherever they intersect.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
. . . . . .
Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx
I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy
I The product is −1, so the tangent lines are perpendicularwherever they intersect.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
. . . . . .
Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx
I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy
I The product is −1, so the tangent lines are perpendicularwherever they intersect.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
. . . . . .
Music Selection
“The Curse of Curves” by Cute is What We Aim ForV63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 24 / 34
. . . . . .
Ideal gases
The ideal gas law relatestemperature, pressure, andvolume of a gas:
PV = nRT
(R is a constant, n is theamount of gas in moles)
..Image credit: Scott Beale / Laughing Squid
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 25 / 34
. . . . . .
Compressibility
DefinitionThe isothermic compressibility of a fluid is defined by
β = −dVdP
1V
with temperature held constant.
Approximately we have
∆V∆P
≈ dVdP
= −βV =⇒ ∆VV
≈ −β∆P
The smaller the β, the “harder” the fluid.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 26 / 34
. . . . . .
Compressibility
DefinitionThe isothermic compressibility of a fluid is defined by
β = −dVdP
1V
with temperature held constant.
Approximately we have
∆V∆P
≈ dVdP
= −βV =⇒ ∆VV
≈ −β∆P
The smaller the β, the “harder” the fluid.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 26 / 34
. . . . . .
Compressibility of an ideal gas
Example
Find the isothermic compressibility of an ideal gas.
SolutionIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then
dPdP
· V+ PdVdP
= 0 =⇒ dVdP
= −VP
Soβ = −1
V· dVdP
=1P
Compressibility and pressure are inversely related.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 27 / 34
. . . . . .
Compressibility of an ideal gas
Example
Find the isothermic compressibility of an ideal gas.
SolutionIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then
dPdP
· V+ PdVdP
= 0 =⇒ dVdP
= −VP
Soβ = −1
V· dVdP
=1P
Compressibility and pressure are inversely related.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 27 / 34
. . . . . .
Nonideal gassesNot that there's anything wrong with that
Example
The van der Waals equationmakes fewer simplifications:(
P+ an2
V2
)(V− nb) = nRT,
where P is the pressure, V thevolume, T the temperature, nthe number of moles of the gas,R a constant, a is a measure ofattraction between particles ofthe gas, and b a measure ofparticle size.
...Oxygen..H
..H
..Oxygen
..H
..H
..Oxygen ..H
..H
.
.
.Hydrogen bonds
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 28 / 34
. . . . . .
Nonideal gassesNot that there's anything wrong with that
Example
The van der Waals equationmakes fewer simplifications:(
P+ an2
V2
)(V− nb) = nRT,
where P is the pressure, V thevolume, T the temperature, nthe number of moles of the gas,R a constant, a is a measure ofattraction between particles ofthe gas, and b a measure ofparticle size.
.
..Wikimedia Commons
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 28 / 34
. . . . . .
Compressibility of a van der Waals gas
Differentiating the van der Waals equation by treating V as a functionof P gives (
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Question
I What if a = b = 0?I Without taking the derivative, what is the sign of
dβdb
?
I Without taking the derivative, what is the sign ofdβda
?
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
. . . . . .
Compressibility of a van der Waals gas
Differentiating the van der Waals equation by treating V as a functionof P gives (
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Question
I What if a = b = 0?I Without taking the derivative, what is the sign of
dβdb
?
I Without taking the derivative, what is the sign ofdβda
?
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
. . . . . .
Compressibility of a van der Waals gas
Differentiating the van der Waals equation by treating V as a functionof P gives (
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Question
I What if a = b = 0?I Without taking the derivative, what is the sign of
dβdb
?
I Without taking the derivative, what is the sign ofdβda
?
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
. . . . . .
Compressibility of a van der Waals gas
Differentiating the van der Waals equation by treating V as a functionof P gives (
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Question
I What if a = b = 0?
I Without taking the derivative, what is the sign ofdβdb
?
I Without taking the derivative, what is the sign ofdβda
?
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
. . . . . .
Compressibility of a van der Waals gas
Differentiating the van der Waals equation by treating V as a functionof P gives (
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Question
I What if a = b = 0?I Without taking the derivative, what is the sign of
dβdb
?
I Without taking the derivative, what is the sign ofdβda
?
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
. . . . . .
Compressibility of a van der Waals gas
Differentiating the van der Waals equation by treating V as a functionof P gives (
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Question
I What if a = b = 0?I Without taking the derivative, what is the sign of
dβdb
?
I Without taking the derivative, what is the sign ofdβda
?
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
. . . . . .
Nasty derivatives
I
dβdb
= −(2abn3 − an2V+ PV3)(nV2)− (nbV2 − V3)(2an3)(2abn3 − an2V+ PV3)2
= −nV3
(an2 + PV2
)(PV3 + an2(2bn− V)
)2 < 0
Idβda
=n2(bn− V)(2bn− V)V2(PV3 + an2(2bn− V)
)2 > 0
(as long as V > 2nb, and it’s probably true that V ≫ 2nb).
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 30 / 34
. . . . . .
Outline
The big idea, by example
ExamplesBasic ExamplesVertical and Horizontal TangentsOrthogonal TrajectoriesChemistry
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 31 / 34
. . . . . .
Using implicit differentiation to find derivatives
Example
Finddydx
if y =√x.
SolutionIf y =
√x, then
y2 = x,
so2y
dydx
= 1 =⇒ dydx
=12y
=1
2√x.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 32 / 34
. . . . . .
Using implicit differentiation to find derivatives
Example
Finddydx
if y =√x.
SolutionIf y =
√x, then
y2 = x,
so2y
dydx
= 1 =⇒ dydx
=12y
=1
2√x.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 32 / 34
. . . . . .
The power rule for rational powers
TheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.First, raise both sides to the qth power:
y = xp/q =⇒ yq = xp
Now, differentiate implicitly:
qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so
xp−1
yq−1 =xp−1
xp−p/q = xp−1−(p−p/q) = xp/q−1
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
. . . . . .
The power rule for rational powers
TheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.First, raise both sides to the qth power:
y = xp/q =⇒ yq = xp
Now, differentiate implicitly:
qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so
xp−1
yq−1 =xp−1
xp−p/q = xp−1−(p−p/q) = xp/q−1
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
. . . . . .
The power rule for rational powers
TheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.First, raise both sides to the qth power:
y = xp/q =⇒ yq = xp
Now, differentiate implicitly:
qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so
xp−1
yq−1 =xp−1
xp−p/q = xp−1−(p−p/q) = xp/q−1
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
. . . . . .
The power rule for rational powers
TheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.First, raise both sides to the qth power:
y = xp/q =⇒ yq = xp
Now, differentiate implicitly:
qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so
xp−1
yq−1 =xp−1
xp−p/q = xp−1−(p−p/q) = xp/q−1
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
. . . . . .
Summary
I Implicit Differentiation allows us to pretend that a relationdescribes a function, since it does, locally, “almost everywhere.”
I The Power Rule was established for powers which are rationalnumbers.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 34 / 34
