Lesson 10: The Chain Rule
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Transcript of Lesson 10: The Chain Rule
. . . . . .
Section2.5TheChainRule
V63.0121.027, CalculusI
October6, 2009
Announcements
I Quiz2thisweekI Midterm inclasson§§1.1–1.4
. . . . . .
CompositionsSeeSection1.2forreview
DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
.
.g .f.x .g(x) .f(g(x))
.f ◦ g
Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.
. . . . . .
CompositionsSeeSection1.2forreview
DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g
.f
.x .g(x)
.f(g(x)).f ◦ g
Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.
. . . . . .
CompositionsSeeSection1.2forreview
DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x)
.f(g(x)).f ◦ g
Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.
. . . . . .
CompositionsSeeSection1.2forreview
DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x) .f(g(x))
.f ◦ g
Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.
. . . . . .
CompositionsSeeSection1.2forreview
DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x) .f(g(x))
.f ◦ g
Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.
. . . . . .
CompositionsSeeSection1.2forreview
DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x) .f(g(x))
.f ◦ g
Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.
. . . . . .
Outline
HeuristicsAnalogyTheLinearCase
Thechainrule
Examples
Relatedratesofchange
. . . . . .
Analogy
Thinkaboutridingabike. Togofasteryoucaneither:
I pedalfasterI changegears
.
.Imagecredit: SpringSun
Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):
φ(θ) =Rθ
r
Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.
. . . . . .
Analogy
Thinkaboutridingabike. Togofasteryoucaneither:
I pedalfaster
I changegears
.
.Imagecredit: SpringSun
Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):
φ(θ) =Rθ
r
Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.
. . . . . .
Analogy
Thinkaboutridingabike. Togofasteryoucaneither:
I pedalfasterI changegears
.
.Imagecredit: SpringSun
Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):
φ(θ) =Rθ
r
Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.
. . . . . .
Analogy
Thinkaboutridingabike. Togofasteryoucaneither:
I pedalfasterI changegears
.
.Imagecredit: SpringSun
Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):
φ(θ) =Rθ
r
Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.
. . . . . .
TheLinearCase
QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)
I ThecompositionisalsolinearI Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.
Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.
. . . . . .
TheLinearCase
QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)
I ThecompositionisalsolinearI Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.
Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.
. . . . . .
TheLinearCase
QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)
I Thecompositionisalsolinear
I Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.
Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.
. . . . . .
TheLinearCase
QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)
I ThecompositionisalsolinearI Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.
Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.
. . . . . .
TheLinearCase
QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)
I ThecompositionisalsolinearI Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.
Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.
. . . . . .
TheNonlinearCaseSeetheMathematicaapplet
Let u = g(x) and y = f(u). Suppose x ischangedbyasmallamount ∆x. Then
∆y ≈ f′(y)∆u
and∆u ≈ g′(u)∆x.
So∆y ≈ f′(y)g′(u)∆x =⇒ ∆y
∆x≈ f′(y)g′(u)
. . . . . .
Outline
HeuristicsAnalogyTheLinearCase
Thechainrule
Examples
Relatedratesofchange
. . . . . .
Theoremoftheday: Thechainrule
TheoremLet f and g befunctions, with g differentiableat x and fdifferentiableat g(x). Then f ◦ g isdifferentiableat x and
(f ◦ g)′(x) = f′(g(x))g′(x)
InLeibniziannotation, let y = f(u) and u = g(x). Then
dydx
=dydu
dudx
..dy
��du��dudx
. . . . . .
Observations
I Succinctly, thederivativeofacompositionistheproductofthederivatives
I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare
I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions
.
.Imagecredit: ooOJasonOoo
. . . . . .
Theoremoftheday: Thechainrule
TheoremLet f and g befunctions, with g differentiableat x and fdifferentiableat g(x). Then f ◦ g isdifferentiableat x and
(f ◦ g)′(x) = f′(g(x))g′(x)
InLeibniziannotation, let y = f(u) and u = g(x). Then
dydx
=dydu
dudx
..dy
��du��dudx
. . . . . .
Observations
I Succinctly, thederivativeofacompositionistheproductofthederivatives
I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare
I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions
.
.Imagecredit: ooOJasonOoo
. . . . . .
CompositionsSeeSection1.2forreview
DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x) .f(g(x))
.f ◦ g
Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.
. . . . . .
Observations
I Succinctly, thederivativeofacompositionistheproductofthederivatives
I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare
I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions .
.Imagecredit: ooOJasonOoo
. . . . . .
Theoremoftheday: Thechainrule
TheoremLet f and g befunctions, with g differentiableat x and fdifferentiableat g(x). Then f ◦ g isdifferentiableat x and
(f ◦ g)′(x) = f′(g(x))g′(x)
InLeibniziannotation, let y = f(u) and u = g(x). Then
dydx
=dydu
dudx
..dy
��du��dudx
. . . . . .
Theoremoftheday: Thechainrule
TheoremLet f and g befunctions, with g differentiableat x and fdifferentiableat g(x). Then f ◦ g isdifferentiableat x and
(f ◦ g)′(x) = f′(g(x))g′(x)
InLeibniziannotation, let y = f(u) and u = g(x). Then
dydx
=dydu
dudx
..dy
��du��dudx
. . . . . .
Outline
HeuristicsAnalogyTheLinearCase
Thechainrule
Examples
Relatedratesofchange
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f(u) =
√u and g(x) = 3x2 + 1. Then
f′(u) = 12u
−1/2, and g′(x) = 6x. So
h′(x) = 12u
−1/2(6x) = 12(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g.
Let f(u) =√u and g(x) = 3x2 + 1. Then
f′(u) = 12u
−1/2, and g′(x) = 6x. So
h′(x) = 12u
−1/2(6x) = 12(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f(u) =
√u and g(x) = 3x2 + 1.
Thenf′(u) = 1
2u−1/2, and g′(x) = 6x. So
h′(x) = 12u
−1/2(6x) = 12(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f(u) =
√u and g(x) = 3x2 + 1. Then
f′(u) = 12u
−1/2, and g′(x) = 6x. So
h′(x) = 12u
−1/2(6x)
= 12(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f(u) =
√u and g(x) = 3x2 + 1. Then
f′(u) = 12u
−1/2, and g′(x) = 6x. So
h′(x) = 12u
−1/2(6x) = 12(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
. . . . . .
Corollary
Corollary(ThePowerRuleCombinedwiththeChainRule)If n isanyrealnumberand u = g(x) isdifferentiable, then
ddx
(un) = nun−1dudx
.
. . . . . .
Doesordermatter?
Example
Findddx
(sin 4x) andcompareittoddx
(4 sin x).
Solution
I Forthefirst, let u = 4x and y = sin(u). Then
dydx
=dydu
· dudx
= cos(u) · 4 = 4 cos 4x.
I Forthesecond, let u = sin x and y = 4u. Then
dydx
=dydu
· dudx
= 4 · cos x
. . . . . .
Doesordermatter?
Example
Findddx
(sin 4x) andcompareittoddx
(4 sin x).
Solution
I Forthefirst, let u = 4x and y = sin(u). Then
dydx
=dydu
· dudx
= cos(u) · 4 = 4 cos 4x.
I Forthesecond, let u = sin x and y = 4u. Then
dydx
=dydu
· dudx
= 4 · cos x
. . . . . .
Doesordermatter?
Example
Findddx
(sin 4x) andcompareittoddx
(4 sin x).
Solution
I Forthefirst, let u = 4x and y = sin(u). Then
dydx
=dydu
· dudx
= cos(u) · 4 = 4 cos 4x.
I Forthesecond, let u = sin x and y = 4u. Then
dydx
=dydu
· dudx
= 4 · cos x
. . . . . .
Ordermatters!
Example
Findddx
(sin 4x) andcompareittoddx
(4 sin x).
Solution
I Forthefirst, let u = 4x and y = sin(u). Then
dydx
=dydu
· dudx
= cos(u) · 4 = 4 cos 4x.
I Forthesecond, let u = sin x and y = 4u. Then
dydx
=dydu
· dudx
= 4 · cos x
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) ddx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) ddx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) ddx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) ddx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) ddx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) ddx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
. . . . . .
A metaphor
Thinkaboutpeelinganonion:
f(x) =
(3√
x5︸︷︷︸�5
−2
︸ ︷︷ ︸3√�
+8
︸ ︷︷ ︸�+8
)2
︸ ︷︷ ︸�2
.
.Imagecredit: photobunny
f′(x) = 2(
3√
x5 − 2 + 8)
13(x
5 − 2)−2/3(5x4)
. . . . . .
Combiningtechniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)
SolutionThe“last”partofthefunctionistheproduct, soweapplytheproductrule. Eachfactor’sderivativerequiresthechainrule:
ddx
((x3 + 1)10 · sin(4x2 − 7)
)=
(ddx
(x3 + 1)10)· sin(4x2−7)+ (x3 +1)10 ·
(ddx
sin(4x2 − 7)
)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
. . . . . .
Combiningtechniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe“last”partofthefunctionistheproduct, soweapplytheproductrule. Eachfactor’sderivativerequiresthechainrule:
ddx
((x3 + 1)10 · sin(4x2 − 7)
)=
(ddx
(x3 + 1)10)· sin(4x2−7)+ (x3 +1)10 ·
(ddx
sin(4x2 − 7)
)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
. . . . . .
Combiningtechniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe“last”partofthefunctionistheproduct, soweapplytheproductrule. Eachfactor’sderivativerequiresthechainrule:
ddx
((x3 + 1)10 · sin(4x2 − 7)
)=
(ddx
(x3 + 1)10)· sin(4x2−7)+ (x3 +1)10 ·
(ddx
sin(4x2 − 7)
)
= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
. . . . . .
Combiningtechniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe“last”partofthefunctionistheproduct, soweapplytheproductrule. Eachfactor’sderivativerequiresthechainrule:
ddx
((x3 + 1)10 · sin(4x2 − 7)
)=
(ddx
(x3 + 1)10)· sin(4x2−7)+ (x3 +1)10 ·
(ddx
sin(4x2 − 7)
)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
. . . . . .
YourTurn
Findderivativesofthesefunctions:
1. y = (1− x2)10
2. y =√sin x
3. y = sin√x
4. y = (2x− 5)4(8x2 − 5)−3
5. F(z) =
√z− 1z + 1
6. y = tan(cos x)
7. y = csc2(sin θ)
8. y = sin(sin(sin(sin(sin(sin(x))))))
. . . . . .
Solutionto#1
ExampleFindthederivativeof y = (1− x2)10.
Solutiony′ = 10(1− x2)9(−2x) = −20x(1− x2)9
. . . . . .
Solutionto#2
ExampleFindthederivativeof y =
√sin x.
SolutionWriting
√sin x as (sin x)1/2, wehave
y′ = 12 (sin x)−1/2 (cos x) =
cos x
2√sin x
. . . . . .
Solutionto#3
ExampleFindthederivativeof y = sin
√x.
Solution
y′ =ddx
sin(x1/2) = cos(x1/2)12x−1/2 =
cos(√
x)
2√x
. . . . . .
Solutionto#4
ExampleFindthederivativeof y = (2x− 5)4(8x2 − 5)−3
SolutionWeneedtousetheproductruleandthechainrule:
y′ = 4(2x− 5)3(2)(8x2 − 5)−3 + (2x− 5)4(−3)(8x2 − 5)−4(16x)
Therestisabitofalgebra, usefulifyouwantedtosolvetheequation y′ = 0:
y′ = 8(2x− 5)3(8x2 − 5)−4 [(8x2 − 5) − 6x(2x− 5)
]= 8(2x− 5)3(8x2 − 5)−4 (
−4x2 + 30x− 5)
= −8(2x− 5)3(8x2 − 5)−4 (4x2 − 30x + 5
)
. . . . . .
Solutionto#5
Example
Findthederivativeof F(z) =
√z− 1z + 1
.
Solution
y′ =12
(z− 1z + 1
)−1/2 ((z + 1)(1) − (z− 1)(1)
(z + 1)2
)=
12
(z + 1z− 1
)1/2 (2
(z + 1)2
)=
1(z + 1)3/2(z− 1)1/2
. . . . . .
Solutionto#6
ExampleFindthederivativeof y = tan(cos x).
Solutiony′ = sec2(cos x) · (− sin x) = − sec2(cos x) sin x
. . . . . .
Solutionto#7
ExampleFindthederivativeof y = csc2(sin θ).
SolutionRememberthenotation:
y = csc2(sin θ) = [csc(sin θ)]2
So
y′ = 2 csc(sin θ) · [− csc(sin θ) cot(sin θ)] · cos(θ)= −2 csc2(sin θ) cot(sin θ) cos θ
. . . . . .
Solutionto#8
ExampleFindthederivativeof y = sin(sin(sin(sin(sin(sin(x)))))).
SolutionRelax! It’sjustabunchofchainrules. Alloftheselinesaremultipliedtogether.
y′ = cos(sin(sin(sin(sin(sin(x))))))
· cos(sin(sin(sin(sin(x)))))
· cos(sin(sin(sin(x))))
· cos(sin(sin(x)))
· cos(sin(x))
· cos(x))
. . . . . .
Outline
HeuristicsAnalogyTheLinearCase
Thechainrule
Examples
Relatedratesofchange
. . . . . .
Relatedratesofchange
QuestionTheareaofacircle, A = πr2,changesasitsradiuschanges. Iftheradiuschangeswithrespecttotime,thechangeinareawithrespecttotimeis
A.dAdr
= 2πr
B.dAdt
= 2πr +drdt
C.dAdt
= 2πrdrdt
D. notenoughinformation
.
.Imagecredit: JimFrazier
. . . . . .
Relatedratesofchange
QuestionTheareaofacircle, A = πr2,changesasitsradiuschanges. Iftheradiuschangeswithrespecttotime,thechangeinareawithrespecttotimeis
A.dAdr
= 2πr
B.dAdt
= 2πr +drdt
C.dAdt
= 2πrdrdt
D. notenoughinformation
.
.Imagecredit: JimFrazier
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Whathavewelearnedtoday?
I Thederivativeofacompositionistheproductofderivatives
I Insymbols:(f ◦ g)′(x) = f′(g(x))g′(x)
I Calculusislikeanonion, andnotbecauseitmakesyoucry!