Partial Derivatives
Chain Rule
Determine if a limit exists:
1.
2.
xy
1*1
= 4
=1/2
x4 + y2
1 + 12
Use two path test to test for non-existence of a limit at a point
First test by substituting,
:
lim (x,y)  (1,1)
Two-Path Test for Nonexistence of a limit: If a function f(x,y) has different limits along two
different paths as (x,y) approaches (x0,y0), then lim (x,y) –> (x0,y0) f(x,y) does not exist.
[Note: most often used when (x0,y0) is the point (0,0). The different paths are defined as
y = mx. By evaluating lim(x,y) –> (0, 0) f(x,mx), if m remains in the result, then the limit varies
along different paths varies as a function of m and therefore does not exist.]
Steps:
•
Set up the generic equation of a line through the point (x0,y0). If (x0,y0) is the point (0,0)
then the equation is y=mx, otherwise the equation is y = m(x-x0) + y0.
•
Substitute for y in the original equation.
•
If m doesn’t disappear, then the limit doesn’t exist because the limit varies with the slope
of the line through the point (x0,y0).
Example:
lim (x,y)  (0,0)
xy
= lim
x4 + y2
(x,mx)(0,0)
x*mx
= lim
x4 + (mx)2
(x,mx)(0,0)
mx2
=
x2(x2 + m2)
1
m
Theorum 5: Chain rule for Functions of Two Independent Variables: If w = f(x,y) has
continuous partial derivatives fx and fy and if x = (t) and y = y(t) are differentiable functions
of t, then the composite function w = f(x(t), y(t)) is a differentiable function of t and
df/dt = fx(x(t), y(t)) * x’(t) + fy(x(t), y(t)) * y’(t) , or
dw df dx df dy
dw dw dx dw dy
also written as
dt = dx dt + dy dt
dt = dx dt + dy dt
Theorum 6: Chain Rule for Functions of Three Independent Variables: If w = f(x,y,z) is
differentiable and x, y, and z are differentiable functions of t, then w is a differentiable
function of t and dw df dx df dy df dz
dt = dx dt + dy dt +dz dt
Theorum 7: Chain Rule for Two Independent Variables and Three Intermediate Variables:
Suppose that w = f(x,y,z), x = g(r,s), y = h(r,s), and z = k(r,s). If all four functions are
differentiable, then w has partial derivatives with respect to r and s given by the formulas
dw dw dx dw dy dw dz
dw dw dx dw dy dw dz
dr = dx dr + dy dr + dz dr
ds = dx ds + dy ds + dz ds
Note the following extensions of Theorum 7: If w = f(x,y), x = g(r,s), and y = h(r,s) then
dw dw dx dw dy
dw dw dx dw dy
dr = dx dr + dy dr
ds = dx ds + dy ds
If w = f(x) and x = g(r,s), then
dw dw dx
dw dw dx
dr = dx dr
ds = dx ds
Examples:
Partial Derivative with Respect to x: The partial derivative of f(x,y) with
df
f(x0+h,y0) – f(x0,y0)
lim
respect to x at the point (x0,y0) is
fx =
=
dx
(x0,y0)
h
Find the derivative of z = f(x,y) = xy along the curve x = t, y = t2 at t = 1
dz
dx
x
provided the limit exists.
Partial Derivative with Respect to y: The partial derivative of f(x,y) with
respect to y at the point (x0,y0) is
df
f(x0,y0+h) – f(x0,y0)
fy = dy
(x0,y0)
=
h
0
dz
dy
y
dx
dt
lim
h
dz/dx = y
z
h
0
dz/dy = x
dx/dt = 1
dy/dt = 2t
dz dz dx dz dy
=
+
= y*1 + x*(2t)
dt
dx dt dy dt
dz
substitutu=ing for x and y
= t2 +(t)(2t) = 3t2
dt
dz
= 3*(1)2 = 3
dt
dy
dt
provided the limit exists.
[Note: In many cases, df/dx  df/dy ]
t
t=1
Notes:
•When calculating df/dx, any y’s in the equation are treated as constants when taking the
derivative. Similarly, when calculating df/dy, any x’s in the equation are treated as constants
when taking the derivative.
•Second order partial derivatives
d2f d df
d2f d df
d2f
d
fxx =
=
, fyy =
=
fxy =
=
dx2 dx dx
dy2 dy dy
dx dy dx
•F(x,y,z) = 0 implicitly defines a function z = G(x,y)
( )
( )
( dfdy)= f
yx =
d2f
d df
=
dy dx dy dx
( )
•Implicit Partial Differentiation, e.g. find dz/dx of xz – y ln z = x + y. Treat y’s like constants.
dxz – dy ln z = dx + dy
dx
dx
dx dx
xdz + zdx – ydln z = dx + dy
dx
dx
dx
dx dx
xdz + z – y dz = dx + 0
dx
z dx dx
dz = 1 – z
dx x – y/z
Suppose f(x,y,z) = x + y
f(x,y,z)
df
dx
x
df
dy
dy
dx du
du
u
df
dz
y
+z2
and x = u/v; y = u + ln(v); and z = u. Find df/du and df/dv
f(x,y,z)
df/dx = 1
df/dy = 1
df
dx
z
df
dy
x
dz
du
dx
dv
df df dx df dy
df dz
=
+
+
du dx du dy du
dz du
dy
dv
v
df
dz
y
df/dz = 2z
z
dz
dv
dx/du = 1/v
dx/dv = – uv –2
dy/du = 1
dy/dv = 1/v
dz/du = 1
dz/dv = 0
df df dx df dy
df dz
=
+
+
dv dx dv dy dv
dz dv
= 1 * 1/v + 1*1 – 2z * 1
= 1 * (–uv –2) + 1*1/v – 2z * 0
= 1/v + 1 – 2z but z = u
= (–uv –2) + 1/v
= 1/v + 1 + –2u