Chain Rule and Implicit Differentiation
Mathematics 54 - Elementary Analysis 2
Chain Rule
Recall:
Supposeyis a differentiable function ofx. Assume further thatxis a differentiable function oft. Thenyis also a differentiable function oft. Moreover,
dy dt =
dy dx·
dx dt.
For functions of several variables, we have the following analogous formulas.
Chain Rule
Letzbe a differentiable function ofxandy. In turn, supposexandyare both differentiable functions oft. Thus,zis also a differentiable function of a single variablet. Moreover,
dz dt =
∂z
∂x· dx dt+
∂z
∂y· dy dt
Chain Rule Diagram
Chain Rule
Ifz=f(x,y), wherex=x(t) andy=y(t), then
dz dt =
∂z
∂x· dx dt+
∂z
∂y· dy dt
Two-Variable Chain Rule
Example
Letz=x2y, wherex=t2andy=t3. Use chain rule to finddz
dt, without direct substitution first.
Solution.We have
dz dt =
∂z
∂x dx dt+
∂z
∂y dy dt
= (2xy)(2t)+(x2)(3t2)
= (2t5)(2t)+(t4)(3t2)
= 7t6
Indeed, by direct calculation,z=x2y=(t2)2(t3)=t7. Thus,dzdt=7t6.
Chain Rule for Several Variables
Chain Rule
Letzbe a differentiable function ofxandy.
In turn, suppose thatxandyare differentiable functions ofsandt. Thus,zis also a differentiable function ofsandt. Moreover,
∂z
∂s =
∂z
∂x
∂x
∂s+
∂z
∂y
∂y
∂s
∂z
∂t =
∂z
∂x
∂x
∂t+
∂z
∂y
∂y
Chain Rule
Example
Letz=exy, wherex=2s+tandy=s
t.
Find∂z
∂s and
∂z
∂t at (s,t)=(−1, 1), using chain rule.
Solution.
∂z
∂s = ye
xy(2)
+xexy
µ1
t
¶
= exy³2y+x t
´
∂z
∂t = ye
xy(1)
+xexy³−s t2
´
= exy³y−x³s t2
´´
If (s,t)=(−1, 1), then (x,y)=(−1,−1). Thus,
∂z
∂s =e
(−1)(−1) µ
2(−1)+−1
1
¶ = −3e
∂z
∂t =e
(−1)(−1) µ
−1−(−1)
µ −1 (1)2
¶¶ = −2e
General Chain Rule
General Chain Rule
Suppose thatuis a differentiable function ofnvariablesx1,x2, . . .,xn. In turn, suppose eachxi,i=1 ton, is a differentiable function ofmvariables t1,t2, . . . ,tm. Thenuis also a differentiable function of themvariables t1, . . . ,tm.
Moreover, for eachj=1 tom, we have
∂u
∂tj =
∂u
∂x1 ∂x1
∂tj +
∂u
∂x2 ∂x2
∂tj + · · · +
∂u
∂xn
∂xn
Chain Rule
Example
Write out the chain rule for the following case:
z=z(x,y), wherex=x(r,s,u,v) andy=y(r,s,u,v)
∂z
∂r =
∂z
∂x
∂x
∂r+
∂z
∂y
∂y
∂r
∂z
∂s =
∂z
∂x
∂x
∂s+
∂z
∂y
∂y
∂s
∂z
∂u =
∂z
∂x
∂x
∂u+
∂z
∂y
∂y
∂u
∂z
∂v =
∂z
∂x
∂x
∂v+
∂z
∂y
∂y
∂v
Chain Rule
Example
Letu=x2+yz, wherex=prcosθ,y=prsinθ,z=p+r.
Findup,uranduθwhenp=2,r=3 andθ=0.
Solution.The appropriate chain rule is as follows:
up = uxxp+uyyp+uzzp
= 2x(rcosθ)+z(rsinθ)+y(1) ur = uxxr+uyyr+uzzr
= 2x(pcosθ)+z(psinθ)+y(1) uθ = uxxθ+uyyθ
= 2x(−prsinθ)+z(prcosθ)
Whenp=2,r=3,θ=0, we have x=2(3) cos 0=6
y=2(3) sin 0=0, z=2+3=5.
Thus, we have
up = 2(6)(3 cos 0)+5(3 sin 0)+0(1)=36
ur = 2(6)(2 cos 0)+5(2 sin 0)+0(1)=24
Chain Rule
Example
Supposez=x2+xywithx=2r−sandy=r2+s. Find∂
2z ∂r2.
Solution.Using chain rule:
∂z
∂r=
∂z
∂x·
∂x
∂r+
∂z
∂y·
∂y
∂r=(2x+y)(2)+(x)(2r)=4x+2y+2xr.
Applying product rule, we get
∂2z ∂r2 =
∂ ∂r
¡
4x+2y+2xr¢
= 4∂x
∂r+2
∂y
∂r+2x
∂r
∂r+2r
∂x
∂r
= 4(2)+2(2r)+2x(1)+2r(2)
Implicit Differentiation
Now considerx3+y3=6xy, whereyis a differentiable function ofx.
Thendydxis obtained by implicit differentation.
That is,
3x2+3y2dy dx=6x
dy dx+6y
Thus,dy dx=
6y−3x2
3y2−6x.
Implicit Differentiation
Supposeyis a differentiable function ofx, and thatF(x,y)=0 gives the implicit relation betweenyandx.
Differentiating with respect tox,
Fx(x,y)
µdx
dx
¶
+Fy(x,y)
µdy
dx
¶ =0.
Thus, with the above assumptions,
Implicit Function Theorem 1
dy dx= −
Fx(x,y) Fy(x,y)
Implicit Function Theorem
Example
Compute fordydxifx3+y3=6xy.
Solution.
First, we writex3+y3=6xyasF(x,y)=0, where
F(x,y)=x3+y3−6xy.
Therefore,
dy
dx = − Fx Fy
= −3x 2
Implicit Differentiation
Consider the equation
x3+y3+z3=1−6xyz
wherezis a differentiable function of two independent variablesxandy.
Then∂∂zxmay be obtained through implicit differentiation.
∂ ∂x
¡
x3+y3+z3¢
= ∂
∂x(1−6xyz) 3x2+0+3z2∂z
∂x = 0−
µ
6yz+6xy∂z
∂x
¶
Thus
¡
3z2+6xy¢∂z
∂x = −3x
2 −6yz
∂z
∂x = −
3x2+6yz 3z2+6xy
Note.∂z
∂y may also be computed similarly, with
∂z
∂y= −
3y2+6xz 3z2+6xy.
Implicit Function Theorem
Whenx,yandz follow an implicit equation, computing for∂∂zxor∂∂zycan be simplified through a multivariate chain rule.
IfF(x,y,z)=0 defines implicitly the relation betweenx,yandz.
Differentiating with respect tox, we get
Fx(x,y,z) dx
dx+Fz(x,y,z)
∂z
∂x=0 With the above assumptions,
Implicit Function Theorem 2
∂z
∂x= − Fx Fz
∂z
∂y = − Fy
Implicit Function Theorem
Example
Letx3+y3+z3=1−6xyz, where z is a function ofxandy.
Find∂z
∂xand
∂z
∂y.
Solution.Let us writex3+y3+z3=1−6xyz as F(x,y,z)=0 where
F(x,y,z)=x3+y3+z3+6xyz−1.
Therefore,
∂z
∂x = −
Fx Fz
= −3x 2+6yz
3z2+6xy ∂z
∂y = −
Fy
Fz
= −3y 2+6xz
3z2+6xy
Exercises
1 Use chain rule to find∂z
∂t forz= x2
py withx=3rtandy=r t.
2 Use chain rule to find∂z
∂t forz=x
2+3xyz+z2withx=2t+1,y=et
andz=3−t2.
3 Use chain rule to findfssforf(x,y)=2x+4xy−y2withx=s+2tand y=tps.
4 Supposef is a differentiable function ofxandy, and
g(u,v)=f(eu+sinv,eu+cosv). Use the following table of values to calculategu(0, 0) andgv(0, 0).
f g fx fy (0, 0) 3 6 4 8 (1, 2) 6 3 2 5