Assignment 4
Practice 25/12/24
Exercise 1
Find all the first-order partial derivatives of the following functions.
1
f (x, y) = e
2
f (x, y) = (x + y)/(xy − 1)
3
f (x, y, z) = (x
−x
sin (x + y)
2
2
+ y
2
−1/2
+ z )
No 1.
∂f
∂
−x
=
∂x
(e
sin(x + y))
∂x
∂
−x
=
(e
−x
∂x
−x
−x
) sin(x + y) + e
cos(x + y)
−x
∂f
( cos(x + y) − sin(x + y))
∂
=
∂y
−x
(e
sin(x + y))
∂y
−x
= e
∂
( sin(x + y))
∂y
−x
= e
cos(x + y)
No 2.
∂f
∂
=
∂x
x + y
(
∂x
)
xy − 1
(xy − 1) ⋅ 1 − (x + y) ⋅ y
=
2
(xy − 1)
xy − 1 − y(x + y)
=
(xy − 1)2
2
xy − 1 − yx − y
=
2
(xy − 1)
2
−1 − y
=
( sin(x + y))
∂x
= (−e
= e
∂
) sin(x + y) + e
(xy − 1)2
∂f
∂
x + y
=
∂y
(
)
∂y
xy − 1
(xy − 1) ⋅ 1 − (x + y) ⋅ x
=
2
(xy − 1)
xy − 1 − x(x + y)
=
2
(xy − 1)
2
xy − 1 − x
=
− xy
(xy − 1)2
2
−1 − x
=
2
(xy − 1)
No 3.
∂f
∂
2
=
((x
∂x
2
+ y
2
−1/2
+ z )
)
∂x
1
= −
2
2
(x
+ y
2
2
−3/2
2
+ z )
⋅ 2x
2
x
= −
(x
∂f
+ y
∂
2
=
((x
∂y
2
3/2
+ z )
2
+ y
2
−1/2
+ z )
)
∂y
1
= −
2
2
(x
+ y
2
2
−3/2
2
+ z )
⋅ 2y
2
y
= −
(x
∂f
+ y
∂
2
=
((x
∂z
2
3/2
+ z )
2
+ y
2
−1/2
+ z )
)
∂z
= −
1
2
2
(x
+ y
2
2
2
−3/2
+ z )
⋅ 2z
2
z
= −
(x
+ y
2
3/2
+ z )
Exercise 2
Given w = e + x ln y + y ln x, verify that w
x
xy
x
w = e
+ x ln y + y ln x
∂
wx =
x
(e
+ x ln y + y ln x)
∂x
x
= e
y
+ ln y +
x
∂
wxy =
=
x
(e
∂y
1
y
+
y
+ ln y +
)
x
1
x
.
= wyx
∂
wy =
x
(e
+ x ln y + y ln x)
∂y
x
=
+ ln x
y
wyx =
∂
(
x
∂x
y
1
1
=
+ ln x)
+
y
x
∴ wxy = wyx
Exercise 3
Find the value of ∂x/∂z and ∂x/∂y at the point (1,-1,-3) from the following equation:
2
xz + y ln x − x
+ 4 = 0
In this case, which variables are considered independent and which are dependent?
y is constant and x is a function of y and z.
∂
2
(xz + y ln x − x
+ 4) = 0
∂z
∂x
x + z
1 ∂x
+ y ⋅
∂z
∂x
− 2x
= 0
x ∂z
∂z
∂
y
(z +
∂z
− 2x) = −x
x
∂x
−x
=
∂z
∂x
z + y/x − 2x
−(1)
−x
=
∂z
=
z + y/x − 2x
(1,−1,−3)
∂
2
(xz + y ln x − x
−3 + (−1/1) − 2(1)
(1,−1,−3)
1
=
6
+ 4) = 0
∂y
∂x
z
1 ∂x
+ ln x + y ⋅
∂y
∂x
− 2x
x ∂y
∂
∂y
(z +
= 0
∂y
y
− 2x) = − ln x
x
∂x
− ln x
=
∂z
∂x
∂z
=
(1,−1,−3)
z + y/x − 2x
−(ln 1)
− ln x
z + y/x − 2x
=
(1,−1,−3)
= 0
−3 + (−1/1) − 2(1)