Mathematics for Engineers and Scientists (MATH1551)
Partial Differentiation
1. Calculate ∂f /∂x and ∂f /∂y when f (x, y) is given by:
p
a) x2 + y 2 sin(xy),
b) (x + y)/(x − y),
c)
e) xy ,
f) log(x2 + y 2 ),
g) xy + x3 cos(xy),
x2 + y 2 ,
d)
p
−1
x2 + y 2
,
h) xy/(x + y).
Solution:
(a) ∂f /∂x = 2x + y 3 cos(xy),
∂f /∂y = 2y sin(xy) + xy 2 cos(xy).
(b) ∂f /∂x = (x − y)−2 ((x − y) − (x + y)) = −2y(x − y)−2 ,
∂f /∂y = (x − y)−2 ((x − y) + (x + y)) = 2x(x − y)−2 .
−1
(c) ∂f /∂x = x (x2 + y 2 ) 2 ,
−1
∂f /∂y = y (x2 + y 2 ) 2 .
−3
(d) ∂f /∂x = −x (x2 + y 2 ) 2 ,
−3
∂f /∂y = −y (x2 + y 2 ) 2 .
(e) ∂f /∂x = yxy−1 ,
∂f /∂y = xy log x,
(to see this, use f (x, y) = xy = ey log x ).
2
2
(f) ∂f /∂x = 2x/ (x + y ),
∂f /∂y = 2y/ (x2 + y 2 ).
(g) ∂f /∂x = y + 3x2 cos(xy) − x3 y sin(xy),
∂f /∂y = x − x4 sin(xy).
(h) ∂f /∂x = y/(x + y) − xy/(x + y)2 ,
∂f /∂y = x/(x + y) − xy/(x + y)2 .
3. Calculate ∂f /∂x, ∂f /∂y, ∂ 2 f /∂x∂y and ∂ 2 f /∂y∂x when f (x, y) is given by
a) x2 y 3 + ex + log y,
b) x2 cos y + x2 + cos y,
d) tan−1 (y/x),
e) (x2 + y 2 )−1/2 .
c) x tan (y 2 ),
Solution:
(a) ∂f /∂x = 2xy 3 + ex , ∂f /∂y = 3x2 y 2 + y1 , ∂ 2 f /∂x∂y = 6xy 2 = ∂ 2 f /∂y∂x.
(b) ∂f /∂x = 2x cos y + 2x, ∂f /∂y = −x2 sin y − sin y, ∂ 2 f /∂x∂y = −2x sin y = ∂ 2 f /∂y∂x.
(c) ∂f /∂x = tan (y 2 ), ∂f /∂y = 2xy sec2 y 2 ,
∂ 2f
∂ ∂f
∂
=
=
2xy sec2 y 2 = 2y sec2 y 2 ,
∂x∂y
∂x ∂y
∂x
∂ 2f
∂ ∂f
∂
=
=
tan y 2 = 2y sec2 y 2 .
∂y∂x
∂y ∂x
∂y
−1
−1
y 2
−2
(d) ∂f /∂x = −yx
1+ x
= −y (x2 + y 2 ) ,
−1
−1
y 2
−1
∂f /∂y = x
1+ x
= x (x2 + y 2 ) ,
1
−2
∂ 2 f /∂x∂y = (x2 + y 2 )
−2
((x2 + y 2 ) (−1) + 2y 2 ) = (y 2 − x2 ) (x2 + y 2 )
−3
− 23
(e) ∂f /∂x = −x (x2 + y 2 ) 2 , ∂f /∂y = −y (x2 + y 2 )
−5
∂ 2 f /∂x∂y = 3xy (x2 + y 2 ) 2 = ∂ 2 f /∂y∂x.
= ∂ 2 f /∂y∂x.
,
6. The two-dimensional Laplace equation is ∂ 2 f /∂x2 + ∂ 2 f /∂y 2 = 0. Find all solutions of the
form ax3 + bx2 y + cxy 2 + dy 3 , with a, b, c and d constant.
Solution:
If f (x, y) = ax3 + bx2 y + cxy 2 + dy 3 , then
∂f
= 3ax2 + 2bxy + cy 2 ,
∂x
∂ 2f
= 6ax + 2by,
∂x2
and so
∂f
= bx2 + 2cxy + 3dy 2 ,
∂y
∂ 2f
= 2cx + 6dy
∂y 2
∂ 2f
∂ 2f
+
= (6a + 2c)x + (2b + 6d)y.
∂x2
∂y 2
Notice that this is true for all x and y. Hence we need 6a + 2c = 0, 2b + 6d = 0 which gives
c = −3a, b = −3d. The most general solution of the required form is
f (x, y) = ax3 − 3dx2 y − 3axy 2 + dy 3 .
7. For the following functions, find df /dt by (i) using the chain rule, and (ii) substituting for
x, y (and z in (c)) to find f (t) explicitly and then differentiating with respect to t:
(a) f (x, y) = x2 + y 2 and x(t) = cos t, y(t) = sin t.
(b) f (x, y) = x2 − y 2 and x(t) = cos t + sin t, y(t) = cos t − sin t.
(c) f (x, y, z) = x/z + y/z and x(t) = cos2 t, y(t) = sin2 t, z(t) = 1/t.
Solution:
(a) (i) We have
df
∂f dx ∂f dy
=
+
dt
∂x dt
∂y dt
= −2x sin t + 2y cos t = 0.
(ii) Now f (t) = cos2 t + sin2 t = 1, so
df
= −2 sin t cos t + 2 cos t sin t = 0.
dt
(b) (i) We have
df
∂f dx ∂f dy
=
+
dt
∂x dt
∂y dt
= 2x (cos t − sin t) − 2y (− sin t − cos t) = 4 cos2 t − sin2 t = 4 cos 2t
(ii) Now f (t) = (cos t + sin t)2 − (cos t − sin t)2 = 4 cos t sin t, so
df
= −4 sin2 t + 4 cos2 t = 4 cos 2t.
dt
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(c) (i) We have
∂f dx ∂f dy ∂f dz
df
=
+
+
dt
∂x dt
∂y dt
∂z dt
2
2
= − cos t sin t + sin t cos t + (x + y)z −2 t−2 = 1.
z
z
(ii) Now f (t) = t cos2 t + t sin2 t = t, so
df
= 1.
dt
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