Partial Differentiation
1
d 2
(a) dx
x = 2x
(b) − sin2 x + cos2 x
(c) cos xesin x
2
(a) 3
(a) fx = ex ey , fy = ex ey , fxx = ex ey , fxy = ex ey fyy = ex ey
(d) fx = − sin x sin y, fy = cos x cos y, fxx = − cos x sin y, fxy = − sin x cos y, fyy =
− cos x sin y
(e) fx = sin y 2 , fy = 2xy cos y 2 , fxx = 0, fxy = 2y cos y 2 fyy = 2x cos y 2 −
4xy 2 sin y 2
(f) fx = ex+sin y , fy = cos y ex+sin y , fxx = ex+sin y , fxy = cos y ex+sin y , fyy =
(cos y − sin y) ex+sin y
4
(a) ∇f = ex ex i + ex ey j
(d) ∇f = (− sin x sin y)i + (cos x
cos y)j
(e) ∇f = (sin y 2 )i + 2xy cos y 2 j
(f) ∇f = (i + cos yj) ex+sin y
5
(a) f (x, y) = 1 + x + y + 21 x2 + 2xy + y 2 + · · ·
(d) f (x, y) = 0 + y + 12 (0) + · · ·
(e) f (x, y) = 0 + 0 + 21 (0 + 0 + 0) + · · · (f) f (x, y) = 1 + x + y + 21 x2 + 2xy + y 2
6
In this answer we write x = x0 + h, y = y0 + k and expand in powers of h and
k.
(a) f (x, y) = ex0 +y0 1 + h + k + 21 h2 + 2hk + k 2 + · · ·
(d) f (x, y) = cos x0 sin y0 +(cos x0 cos y0 )k+(− sin x0 sin y0 )h− 12 cos x0 sin y0 h2 + k 2 −
sin x0 cos y0 hk + · · · (e) f (x, y) = x0 sin y02 + sin y02 h + 2y0 x0 cos y02 k
2
2
+ 2y0 cos y02 hk + xh0 cos y02 − 2x0 y02 sin
y0 k + · · ·
i
(f) f (x, y) = ex0 +sin y0 1 + h + cos y0 k +
7+8
(a) H f =
(d) H f =
x+y
e
ex+y
h2
2
+ cos y0 hk +
ex+y
.
ex+y
− sin x cos y
.
− cos x sin y
− cos x sin y
− sin x cos y
1
− sin y0 2
k
2
(e) H f =
(f) H f =
2ycos y 2
.
2x cos y 2 − 4xy 2 sin y 2
0
2y cos y 2
ex+sin y
cos y ex+sin y
cos yex+sin y
.
(− sin y + cos y) ex+sin y
Chain rule & stationary points
1
(a) fx = 1 + 2x, fy = 2y, fxx = 2, fxy = 0, fyy = 2.
2
Stationary point: fx = fy = 0 =⇒ (x, y) = (−1/2, 0). Here ∆ = fxx fyy −fxy
=
4 > 0. Also fx x = 2 > 0 so this point is a minimum.
(b) fx = 2x, fy = 1 − 2y, fxx = 2, fxy = 0, fyy = −2.
2
=
Stationary point: fx = fy = 0 =⇒ (x, y) = (0, 1/2). Here ∆ = fxx fyy − fxy
−4 < 0. Hence this pooint is a saddle.
(c) Similar method gives maximum at (1, 1).
(d) Maximum at (1/3, 1/3).
(e) 5 points: (i) max at (0, 0), (ii) saddle at (0, 1), (iii) saddle at (0, −1), (iv)
saddle at (1, 0), (v) saddle at (−1, 0).
2
∇f = (2x, −2y).
√
(a) Directional derivative
is (2,
√
√−2) · (1, 1)/ 2 = 0.
(b) (−2, −2) · (1, 1)/ 2 = −2 2.
(c) (0, −2) · (0, −1) = 2.
(d) (2, 0) · (0, 1) = 0.
3
Two options:
Method 1 Substitute for x andy into f to get f (t) = . . . and then differentiate.
∂f dx
∂f dy
Method 2 Use the formulate df
dt = ∂x dt + ∂y dt .
(a)
(b)
(c)
(d)
df
dt
df
dt
df
dt
df
dt
=
∂f dx
∂x dt
+
∂f dy
∂y dt
= y(− sin t) + x(cos t) = − sin2 t + cos2 t.
= 0.
= 3t2 + 2t + 1.
= −2t−3 .
4 The Jacobian matrix is given by
∂x ∂y 1
∂u
J = ∂u
=
∂y
∂x
−1
∂v
∂v
Then
∂f ∂u
∂f
∂v
=J
2
∂f
∂x
∂f
∂y
2u
−2v
!
.
(a) If f = x2 − y, we have
∂f
∂x
= 2x, ∂f
∂y = −1, and so
∂f
∂u
= 2x − 2u =
∂f
∂v
= −2x + 2v = −2u + 4v.
2(u − v) − 2u = −2v,
∂f
2 ∂f
2
(b) If f = xy 2 , we have ∂f
∂x = y , ∂y = 2xy, and so ∂u = y + 4xyu,
−y 2 − 4xyv.
∂f
y
y
(c) If f = ey sin x, we have ∂f
∂x = e cos x, ∂y = e sin x, and so
∂f
∂y
=
2
2
∂f
= ey cos x + 2uey sin x = eu −v [cos(u − v) + 2u sin(u − v)] ,
∂u
2
2
∂f
= −ey cos x + 2uey sin x = −eu −v [cos(u − v) − 2v sin(u − v)] .
∂v
∂f
(d) If f = sin x sin y, we have ∂f
∂x = cos x sin y, ∂y = sin x cos y, and so
∂f
= cos x sin y+2u sin x cos y = cos(u−v) sin u2 − v 2 +2u sin(u−v) cos u2 − v 2 ,
∂u
∂f
= − cos x sin y−2v sin x cos y = − cos(u−v) sin u2 − v 2 −2v sin(u−v) cos u2 − v 2 .
∂v
5
(a)
∂f
= (cos θ + sin θ) er(cos θ+sin θ) ,
∂r
∂f
= r (− sin θ + cos θ) er(cos θ+sin θ) .
∂θ
(1)
(2)
(d)
∂f
= − cos θ sin (r cos θ) sin (r sin θ) + sin θ cos (r cos θ) cos (r sin θ) ,
∂r
∂f
= r sin θ sin (r sin θ) sin (r sin θ) + r cos θ cos (r cos θ) cos (r sin θ) .
∂θ
(3)
(4)
(e)
∂f
= cos θ sin r2 sin2 θ + sin θ2r cos θr sin θ cos r2 sin2 θ ,
∂r
∂f
= −r sin θ sin r2 sin2 θ + r cos θ2r cos θr sin θ cos r2 sin2 θ .
∂θ
(5)
(6)
(f)
∂f
= cos θer cos θ+sin(r sin θ) + sin θ cos (r sin θ) er cos θ+sin(r sin θ) ,
∂r
∂f
= −r sin θer cos θ+sin(r sin θ) + r cos θ cos (r sin θ) er cos θ+sin(r sin θ) .
∂θ
3
(7)
(8)