Solutions to Examples on Partial Derivatives
∂f
∂f
= 3;
= 4.
∂x
∂y
∂f
∂f
f (x, y) = xy 3 + x2 y 2 ;
= y 3 + 2xy 2 ;
= 3xy 2 + 2x2 y.
∂x
∂y
∂f
∂f
f (x, y) = x3 y + ex ;
= 3x2 y + ex ;
= x3 .
∂x
∂y
∂f
∂f
f (x, y) = xe2x+3y ;
= 2xe2x+3y + e2x+3y ;
= 3xe2x+3y .
∂x
∂y
x−y
.
f (x, y) =
x+y
∂f
x + y − (x − y)
2y
=
=
;
∂x
(x + y)2
(x + y)2
∂f
−(x + y) − (x − y)
2x
=
=−
.
2
∂y
(x + y)
(x + y)2
1. (a) f (x, y) = 3x + 4y;
(b)
(c)
(d)
(e)
(f) f (x, y) = 2x sin(x2 y).
∂f
= 2x . cos(x2 y) . 2xy + 2 sin(x2 y) = 4x2 y cos(x2 y) + 2 sin(x2 y);
∂x
∂f
= 2x . cos(x2 y) . x2 = 2x3 cos(x2 y).
∂y
2.
f (x, y, z) = x cos z + x2 y 3 ez .
∂f
= cos z + 2xy 3 ez ,
∂x
∂f
= 3x2 y 2 ez ,
∂y
∂f
= −x sin z + x2 y 3 ez .
∂z
1
3. (i) f (x, y) = x2 sin y + y 2 cos x.
fx = 2x sin y − y 2 sin x;
fy = x2 cos y + 2y cos x.
fxx = 2 sin y − y 2 cos x;
fyy = −x2 sin y + 2 cos x;
fxy = 2x cos y − 2y sin x;
fyx = 2x cos y − 2y sin x.
So fxy = fyx .
y
ln x.
x
y 1
y
y
1
fx = . − 2 ln x = 2 (1 − ln x); fy = ln x.
x x x x
x
y
1
2y
y
fxx = 2 . −
− 3 (1 − ln x) = 3 (2 ln x − 3); fyy = 0.
x
x
x
x
1
1 1
1
1
fxy = 2 (1 − ln x); fyx = . − 2 ln x = 2 (1 − ln x).
x
x x x
x
(ii) f (x, y) =
So fxy = fyx .
4.
1
.
+ y2
∂f
∂f ∂x ∂f ∂y
=
+
.
∂t
∂x ∂t
∂y ∂t
∂f
−2x
−2r cos t ∂x
= 2
=
;
= −r sin t;
2
2
∂x
(x + y )
r4
∂t
∂f
−2y
−2r sin t ∂y
= 2
=
;
= r cos t.
2
2
∂y
(x + y )
r4
∂t
∂f
2r2 sin t cos t 2r2 sin t cos t
Hence,
=
−
= 0.
∂t
r4
r4
∂f
∂f ∂x ∂f ∂y
=
+
.
∂r
∂x ∂r
∂y ∂r
∂f
∂f
∂x
∂y
and
are as above.
= cos t,
= sin t
∂x
∂y
∂r
∂r
∂f
−2r cos2 t − 2r sin2 t
−2(cos2 t + sin2 t)
−2
Hence,
=
=
= 3.
∂r
r4
r3
r
f (x, y) =
x2
2
5.
f (x, y) = x2 + xy − y 2 .
(i) f (r, θ) = (r cos θ)2 + (r cos θ) (r sin θ) − (r sin θ)2
= r2 (cos2 θ + cos θ sin θ − sin2 θ).
∂f
= 2r (cos2 θ + cos θ sin θ − sin2 θ).
∂r
∂f
= r2 (−2 cos θ sin θ + cos θ cos θ − sin θ sin θ − 2 sin θ cos θ)
∂θ
= r2 (cos2 θ − sin2 θ − 4 cos θ sin θ).
(ii)
∂x
= cos θ;
∂r
∂x
∂y
∂y
= −r sin θ;
= sin θ;
= r cos θ.
∂θ
∂r
∂θ
∂f
∂f ∂x ∂f ∂y
By the chain rule
=
+
= (2x + y) cos θ + (x − 2y) sin θ
∂r
∂x ∂r ∂y ∂r
= (2r cos θ + r sin θ) cos θ + (r cos θ − 2r sin θ) sin θ
= 2r(cos2 θ + cos θ sin θ − sin2 θ).
∂f
∂f ∂x ∂f ∂y
=
+
= (2x + y) (−r sin θ) + (x − 2y) r cos θ
∂θ
∂x ∂θ
∂y ∂θ
= (2r cos θ + r sin θ) (−r sin θ) + (r cos θ − 2r sin θ) r cos θ
= r2 (cos2 θ − sin2 θ − 4 cos θ sin θ).
6.
f (x, y) = x3 y − y 3 x;
∂f
∂f ∂x ∂f ∂y
=
+
;
∂u
∂x ∂u ∂y ∂u
u
.
v
∂f
∂f ∂x ∂f ∂y
=
+
.
∂v
∂x ∂v
∂y ∂v
x = uv;
y=
∂f
u u3
u3
= 3x2 y − y 3 = 3u2 v 2 − 3 = 3u3 v − 3 .
∂x
v
v
v
3
2
3
u
u
3u
∂f
= x3 − 3y 2 x = 3 − 3 2 uv = u3 v 3 −
.
∂y
v
v
v
∂x
∂x
∂y
1
∂y
−u
= v;
= u;
= ;
= 2.
∂u
∂v
∂u
v
∂v
v
!
!
3u3
u3
3u v − 3 v + u3 v 3 −
v
v
3
3
u
3u
= 3u3 v 2 − 2 + u3 v 2 − 2 = 4u3 v 2 −
v
v
∂f
=
∂u
3
3
1
v
4u3
.
v2
!
!
u3
3u3 −u
3u v − 3 u + u3 v 3 −
v
v
v2
u4
3u4
2u4
= 3u4 v − 3 − u4 v + 3 = 2u4 v + 3 .
v
v
v
∂f
=
∂v
3
7. f (x, y, z) = 2y − sin(xz), x = 3t, y = et−1 , z = ln t.
∂f
∂t
∂f
∂x
dx
dt
∂f
∂t
∂f dx ∂f dy
∂f dz
+
+
.
∂x dt
∂y dt
∂z dt
∂f
∂f
= −z cos(xz);
= 2;
= −x cos(xz).
∂y
∂z
dy
dz
1
= 3;
= et−1 ;
= .
dt
dt
t
x cos(xz)
= −3z cos(xz) + 2et−1 −
t
3t cos(3t ln t)
= −3 ln t cos(3t ln t) + 2et−1 −
t
=
= −3 cos(3t ln t)(1 + ln t) + 2et−1 .
8. f (x, y) = x2 + xy + y 2 , x = uv, y = u/v.
To show that ufu + vfv = 2xfx and ufu − vfv = 2yfy
we need to find fu , fv , fx and fy .
∂f
∂f ∂x ∂f ∂y
=
+
;
∂u
∂x ∂u ∂y ∂u
∂f
∂f ∂x ∂f ∂y
=
+
.
∂v
∂x ∂v
∂y ∂v
1
2u
fu = (2x + y)(v) + (x + 2y)( ) = 2uv 2 + 2u + 2 .
v
v
2
2u
−u
fv = (2x + y)(u) + (x + 2y)( 2 ) = 2u2 v − 3 .
v
v
u
fx = 2x + y = 2uv + . So, 2xfx = 2uvfx = 4u2 v 2 + 2u2 .
v
2u
4u2
2u
fy = x + 2y = uv +
. So, 2yfy =
fx = 2u2 + 2 .
v
v
v
2
2
2u
2u
Now ufu + vfv = 2u2 v 2 + 2u2 + 2 + 2u2 v 2 − 2
v
v
= 4u2 v 2 + 2u2 = 2xfx as required,
fu =
fv =
4
and ufu − vfv = 2u2 v 2 + 2u2 +
= 2u2 +
2u2
2u2
2 2
−
2u
v
+
v2
v2
4u2
= 2yfy as required.
v2
9. u(x, y) = ln(1 + xy 2 ).
∂u
1
y2
∂2u
−y 2 . y 2
y4
2
=
.
y
=
;
=
=
−
.
∂x
1 + xy 2
1 + xy 2
∂x2
(1 + xy 2 )2
(1 + xy 2 )2
∂2u
(1 + xy 2 ) . 2y − 2xy . y 2
2y
=
=
.
∂y∂x
(1 + xy 2 )2
(1 + xy 2 )2
∂2u
∂2u
2y 4
2y 4
Hence 2 2 + y 3
=−
+
= 0.
∂x
∂y∂x
(1 + xy 2 )2 (1 + xy 2 )2
10. u(x, y) = x2 cosh(xy 2 + 1).
NOTE.
d
d
(sinh x) = cosh x; (cosh x) = sinh x.
dx
dx
∂u
= x2 sinh(xy 2 + 1) . y 2 + 2x cosh(xy 2 + 1)
∂x
= x2 y 2 sinh(xy 2 + 1) + 2x cosh(xy 2 + 1)
∂u
= x2 sinh(xy 2 + 1) . 2xy = 2x3 y sinh(xy 2 + 1)
∂y
Hence
2x
∂u
∂u
−y
= 2x3 y 2 sinh(xy 2 +1)+4x2 cosh(xy 2 +1)−2x3 y 2 sinh(xy 2 +1)
∂x
∂y
= 4x2 cosh(xy 2 + 1) = 4u.
11.
Hence
1
∂w
=
2c
∂t
2x + 2ct
∂2w
−4c2
=
∂t2
(2x + 2ct)2
∂w
1
=
2
∂x
2x + 2ct
∂2w
−4
=
2
∂x
(2x + 2ct)2
2
∂2w
2∂ w
=
c
.
∂t2
∂x2
5