chain rule in partial differentiation
DESCRIPTION
Introducing Chain Rule with examplesTRANSCRIPT
PARTIAL DIFFERENTIATION
SEBASTIAN VATTAMATTAM
1. Partial Derivatives and Chain Rules
Definition 1.1. Let f be a function of several variables.Its derivative with respect to one of those variables, keepingothers constant, is called a partial derivative.
The partial derivative of function f with respect to x isdenoted by ∂f
∂x or fx.
Definition 1.2. Let f be a function of x, y.∂f∂x ,
∂f∂y are called the first order partial derivatives of f.
See Figure 1The second order partial derivatives are
∂2f
∂x2=
∂
∂x(∂f
∂x)
∂2f
∂y2=
∂
∂y(∂f
∂y)
∂2f
∂x∂y=
∂
∂x(∂f
∂y)
∂2f
∂y∂x=
∂
∂y(∂f
∂x)
Example 1.3.
f(x, y) = x2 sin y1
2 SEBASTIAN VATTAMATTAM
Figure 1. First Order Partial Derivatives
∂f
∂x= 2x sin y
∂f
∂y= x2 cos y
∂2f
∂x2= 2 sin y
∂2f
∂x∂y= 2x cos y
∂2f
∂y∂x= 2x cos y
∂2f
∂y2= −x2 sin y
PARTIAL DIFFERENTIATION 3
Theorem 1.4.∂2f
∂x∂y=
∂2f
∂y∂x
Theorem 1.5. Chain Rule 1If z = f(x, y) and x = x(s, t), y = y(s, t) then
∂z
∂s=
∂z
∂x
∂x
∂s+
∂z
∂y
∂y
∂s∂z
∂t=
∂z
∂x
∂x
∂t+
∂z
∂y
∂y
∂t
See Figure 2
Figure 2. Chain Rule 1
4 SEBASTIAN VATTAMATTAM
Example 1.6.
z = log(u2 + v);u = ex+y2, v = x2 + y
∂z
∂u=
2u
u2 + v∂z
∂v=
1
u2 + v∂u
∂x= ex+y2
∂u
∂y= 2yex+y2
∂v
∂x= 2x
∂v
∂y= 1
∂z
∂x=
∂z
∂u
∂u
∂x+∂z
∂v
∂v
∂x=
2u
u2 + vex+y2+
1
u2 + v2x =
2
u2 + v(uex+y2+x)
∂z
∂y=
∂z
∂u
∂u
∂y+∂z
∂v
∂v
∂y=
2u
u2 + v2yex+y2+
1
u2 + v=
1
u2 + v(4uyex+y2+1)
Theorem 1.7. Chain Rule 2If z = f(x, y) and x = x(t), y = y(t) then
dz
dt=
∂z
∂x
dx
dt+
∂z
∂y
dy
dt
See Figure 3
Example 1.8.
u = sin(x/y);x = et, y = t2
PARTIAL DIFFERENTIATION 5
Figure 3. Chain Rule 2
∂u
∂x=
1
ycos(x/y)
∂u
∂y=−xy2
cos(x/y)
dx
dt= et
dy
dt= 2t
du
dt=
∂u
∂x
dx
dt+
∂u
∂y
dy
dt=
1
ycos(x/y)et +
−xy2
cos(x/y)2t
Substituting x = et, y = t2, and simplifying we get
du
dt=
t− 2
t3cos(
et
t2)
Theorem 1.9. Chain Rule 3
6 SEBASTIAN VATTAMATTAM
If z = f(x, y) and y = g(x) then
dz
dx=
∂z
∂x+
∂z
∂y
dy
dx
See Figure 4
Figure 4. Chain Rule 3
Example 1.10.
z = x2 +√y; y = sinx
∂z
∂x= 2x
∂z
∂y=
1
2√y
dy
dx= cos x
dz
dx=
∂z
∂x+
∂z
∂y
dy
dx= 2x +
1
2√y
cosx = 2x +cosx
2√
sinx
Mary Bhavan, EttumanoorE-mail address: [email protected]