Solutions to Examples on Partial Derivatives - George Mason University
[Pages:5]Solutions to Examples on Partial Derivatives
f
f
1. (a) f (x, y) = 3x + 4y;
= 3;
= 4.
x
y
(b) f (x, y) = xy3 + x2y2; f = y3 + 2xy2; f = 3xy2 + 2x2y.
x
y
(c) f (x, y) = x3y + ex; f = 3x2y + ex; f = x3.
x
y
(d) f (x, y) = xe2x+3y; f = 2xe2x+3y + e2x+3y; f = 3xe2x+3y.
x
y
(e) f (x, y) = x - y . x+y
f x + y - (x - y)
2y
= x
(x + y)2
= (x + y)2 ;
f -(x + y) - (x - y)
2x
= y
(x + y)2
= - (x + y)2 .
(f) f (x, y) = 2x sin(x2y).
f = 2x . cos(x2y) . 2xy + 2 sin(x2y) = 4x2y cos(x2y) + 2 sin(x2y); x f = 2x . cos(x2y) . x2 = 2x3 cos(x2y). y
2. f (x, y, z) = x cos z + x2y3ez.
f = cos z + 2xy3ez, x f = 3x2y2ez, y f = -x sin z + x2y3ez. z
1
3. (i) f (x, y) = x2 sin y + y2 cos x.
fx = 2x sin y - y2 sin x; fy = x2 cos y + 2y cos x.
fxx = 2 sin y - y2 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
(ii) f (x, y) =
ln x.
x
y1 y
y
1
fx
=
x
. x
-
x2
ln x
=
x2 (1 - ln x);
fy = x ln x.
y
1 2y
y
fxx = x2 .
- x
- x3 (1 - ln x) = x3 (2 ln x - 3);
fyy = 0.
1
11 1
1
fxy = x2 (1 - ln x); fyx = x . x - x2 ln x = x2 (1 - ln x).
So fxy = fyx.
1 4. f (x, y) = x2 + y2 .
f f x f y
=
+
.
t x t y t
f
-2x
-2r cos t x
x = (x2 + y2)2 = r4 ; t = -r sin t;
f
-2y
-2r sin t y
y = (x2 + y2)2 =
r4
; = r cos t. t
f 2r2 sin t cos t 2r2 sin t cos t
Hence, = t
r4
-
r4
= 0.
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, = r
r4
=
r3
= r3 .
2
5. f (x, y) = x2 + xy - y2. (i) f (r, ) = (r cos )2 + (r cos ) (r sin ) - (r sin )2
= r2(cos2 + cos sin - sin2 ).
f = 2r (cos2 + cos sin - sin2 ). r f = r2 (-2 cos sin + cos cos - sin sin - 2 sin cos ) = r2 (cos2 - sin2 - 4 cos sin ).
x
x
y
y
(ii) = cos ;
= -r sin ;
= sin ;
= r cos .
r
r
f f x f y
By the chain rule =
+
= (2x + y) cos + (x - 2y) sin
r x r y r
= (2r cos + r sin ) cos + (r cos - 2r sin ) sin
= 2r(cos2 + cos sin - sin2 ). f f x f y = x + y = (2x + y) (-r sin ) + (x - 2y) r cos = (2r cos + r sin ) (-r sin ) + (r cos - 2r sin ) r cos
= r2 (cos2 - sin2 - 4 cos sin ).
6. f (x, y) = x3y - y3x;
x = uv;
u y= .
v
f f x f y
=
+
;
u x u y u
f f x f y
=
+
.
v x v y v
f x
=
3x2y
- y3
=
3u2v2 u v
-
u3 v3
=
3u3v
-
u3 v3 .
f y
=
x3 - 3y2x
=
u3 v3
u2 - 3 v2 uv
= u3v3 -
3u3 .
v
x
x
y 1 y -u
= v; u
= u; v
=; u v
v = v2 .
f =
u
3u3v
-
u3 v3
v+
u3v3 - 3u3 v
1 v
=
3u3v2
-
u3 v2
+ u3v2
-
3u3 v2
=
4u3v2
-
4u3 v2 .
3
f =
v
3u3v
-
u3 v3
u + u3v3 - 3u3 v
-u v2
=
3u4v
-
u4 v3
-
u4v
+
3u4 v3
=
2u4v
+
2u4 v3 .
7. f (x, y, z) = 2y - sin(xz), x = 3t, y = et-1, z = ln t.
f f dx f dy f dz
=
+
+
.
t x dt y dt z dt
f
f
f
= -z cos(xz);
= 2;
= -x cos(xz).
x
y
z
dx = 3;
dy = et-1;
dz 1 =.
dt
dt
dt t
f = -3z cos(xz) + 2et-1 - x cos(xz)
t
t
= -3 ln t cos(3t ln t) + 2et-1 - 3t cos(3t ln t) t
= -3 cos(3t ln t)(1 + ln t) + 2et-1.
8. f (x, y) = x2 + xy + y2, 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
f f x f y
fu
=
u
=
x
u
+
y
; u
fv
=
v
=
x
v
+
y
. v
fu
=
(2x
+
y)(v)
+
(x
+
1 2y)( )
v
=
2uv2
+
2u
+
2u v2 .
fv fx
= =
(2x + y)(u) + (x + u
2x + y = 2uv + . v
-u 2y)( v2 ) So, 2xfx
= =
2u2v
-
2u2 v3
.
2uvfx = 4u2v2
+
2u2.
fy
=
x + 2y
=
uv
+
2u .
v
So,
2yfy
=
2u v fx
=
2u2
+
4u2 v2 .
Now
ufu
+
vfv
=
2u2v2
+
2u2
+
2u2 v2
+
2u2v2
-
2u2 v2
= 4u2v2 + 2u2 = 2xfx as required,
4
and
ufu
-
vfv
=
2u2v2
+
2u2
+
2u2 v2
-
2u2v2
+
2u2 v2
=
2u2
+
4u2 v2
=
2yfy
as
required.
9. u(x, y) = ln(1 + xy2).
u x
=
1
1 + xy2
. y2
=
1
y2 + xy2 ;
2u -y2 . y2
y4
x2 = (1 + xy2)2 = - (1 + xy2)2 .
2u (1 + xy2) . 2y - 2xy . y2
2y
= yx
(1 + xy2)2
= (1 + xy2)2 .
Hence
2u 2 x2
+ y3 2u yx
=
2y4 - (1 + xy2)2
+
2y4 (1 + xy2)2
=
0.
10. u(x, y) = x2 cosh(xy2 + 1).
d
d
NOTE. (sinh x) = cosh x; (cosh x) = sinh x.
dx
dx
u = x2 sinh(xy2 + 1) . y2 + 2x cosh(xy2 + 1) x
= x2y2 sinh(xy2 + 1) + 2x cosh(xy2 + 1)
u = x2 sinh(xy2 + 1) . 2xy = 2x3y sinh(xy2 + 1) y Hence
u u 2x -y
=
2x3y2
sinh(xy2+1)+4x2
cosh(xy2+1)-2x3y2
sinh(xy2+1)
x y
= 4x2 cosh(xy2 + 1) = 4u.
11.
w
1
=
2c
t 2x + 2ct
w
1
=
2
x 2x + 2ct
Hence
2w t2
=
c2
2w x2
.
2w
-4c2
t2 = (2x + 2ct)2
2w
-4
x2 = (2x + 2ct)2
5
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