HOMEWORK 4 FOR MATH 2250 - MODEL SOLUTION
• When you write your calculation of derivatives, please write the step “plugging your functions for differentiation rules”. I wrote the step in blue color. It
guarantees partial score if your final answer is incorrect.
• Do not copy others’ homework.
Every problem will be worth two points.
(1) Find the first and second derivatives.
y = 4 − 2x − x−3 .
dy
= −2 − (−3)x−4 = −2 + 3x−4 .
dx
d2 y
= 3(−4)x−5 = −12x−5 .
dx2
(2) Find the derivative by (a) applying the Product Rule and (b) by multiplying the
factors to produce a sum of simpler terms to differentiate.
y = (2x + 3)(5x2 − 4x).
(a) Applying the Product Rule
dy
=
dx
=
=
=
d
d
(5x2 − 4x) + (5x2 − 4x) (2x + 3)
dx
dx
2
(2x + 3)(10x − 4) + (5x − 4x)2
20x2 − 8x + 30x − 12 + 10x2 − 8x
30x2 + 14x − 12.
(2x + 3)
(b) Multiplying the factors
y = (2x + 3)(5x2 − 4x) = 10x3 − 8x2 + 15x2 − 12x = 10x3 + 7x2 − 12x.
dy
= 30x2 + 14x − 12.
dx
(3) Find the first and second derivatives of
w = 3z 2 e2z .
d 2z
d
d
d
(e ) = (ez · ez ) = ez ez + ez ez = ez · ez + ez · ez = 2e2z .
dz
dz
dz
dz
dw
d
d
= 3z 2 e2z + e2z (3z 2 ) = 3z 2 2e2z + e2z 6z = (6z 2 + 6z)e2z .
dz
dz
dz
Date: February 11, 2012.
1
2
HOMEWORK 4 FOR MATH 2250 - MODEL SOLUTION
d2 w
d 2z
2z d
2
(e
)
+
e
(6z 2 + 6z) = (6z 2 + 6z)2e2z + e2z (12z + 6)
=
(6z
+
6z)
dz 2
dz
dz
= (12z 2 + 12z + 12z + 6)e2z = (12z 2 + 24z + 6)e2z .
If you have some complicated parts (in this example, e2z ), it is a good idea to
compute the derivative of it separately.
(4) Suppose u and v are differentiable functions of x and that
u(1) = 2,
u0 (1) = 0,
v(1) = 5,
v 0 (1) = −1.
Find the values of the following derivatives at x = 1.
d
(a)
(uv).
dx
d
d
d
(uv) = u v + v u.
dx
dx
dx
d
(uv)
= u(1)v 0 (1) + v(1)u0 (1) = 2 · (−1) + 5 · 0 = −2.
dx
x=1
d u
.
(b)
dx v
d
d
u − u dx
v
d u v dx
=
.
2
dx v
v
d u v(1)u0 (1) − u(1)v 0 (1)
5 · 0 − 2 · (−1)
2
=
=
=
.
dx v x=1
v(1)2
52
25
d v
(c)
.
dx u
d
d
v − v dx
u
d v u dx
=
.
2
dx u
u
u(1)v 0 (1) − v(1)u0 (1)
2 · (−1) − 5 · 0
−2
1
d v =
=
=
=
−
.
dx u x=1
u(1)2
22
4
2
d
(d)
(7v − 2u).
dx
d
dv du (7v − 2u)
=7 −2 = 7v 0 (1) − 2u0 (1) = 7 · (−1) − 2 · 0 = −7.
dx
dx x=1
dx x=1
x=1