Math 1310: CSM
MORE sample exam 2 problems
SOLUTIONS
PLEASE NOTE that the second exam will cover all material from Chapters 2 and 3 (specifically:
Sections 2.1, 2.2, 3.1, 3.2, 3.3, 3.5, 3.6, and 3.7), and the corresponding tutorials.
There will NO be any programming questions on the exam. You certainly may want to use
your calculator for some computations, though.
1. Find (and SHOW ALL YOUR WORK):
(a) y 0 if y = cos(x) + 4x2 5 tan(x). y 0 = sin(x) + 8x 5 sec2 (x)
d
4
(b)
(4z 5 z 3 2z + 1)5 . 5 20z 4 3z 2 2 4z 5 z 3 2z + 1
dz
(c) f 0 (x) if f (x) = sin(x3 4x + 5). f 0 (x) = 3x2 4 cos x3 4x + 5
(d) g 0 (t) if g(t) = tan(t2 2) + t2 2. g 0 (t) = 2t sec2 t2 2 + 2t
(e) h0 (x) if h(x) = sin(sin(sin(x))). h0 (x) = cos(x) cos(sin(x)) cos(sin(sin(x)))
@z
@z
@z
(f)
and
if z = x3 y xy 3 + cos(x2 sin(y)).
= 3x2 y y 3 2x sin(y) sin x2 sin(y) ;
@x
@y
@x
@z
= x3 3xy 2 x2 cos(y) sin x2 sin(y)
@y
(g) q 0 (1) if q(x) = F (x2 + x) and F 0 (2) = 5. q 0 (1) = 15
q
p
p
(h) k0 (x) if k(x) = 4 14 + 3 1 + x.
1
k0 (x) =
⇣ pp
⌘3/4 p
2/3 p
24 3 x + 1 + 14
( x + 1)
x
2. Let
f (x, y) = 3x2 + xy
(a) Find fx (x, y) and fy (x, y). fx (x, y) = 6x
xy 2 .
y 2 + y; fy (x, y) = x
2xy.
(b) Find
@
@
fx (x, y) and
fy (x, y).
@y
@x
What do you notice about these two “mixed partial derivatives”?
@
@
fx (x, y) =
(6x
@y
@y
y 2 + y) =
2y + 1;
@
@
fy (x, y) =
(x
@x
@x
2xy) = 1
2y.
The mixed partial derivatives are the same!!
3. Using the facts that sec(x) = 1/ cos(x), tan(x) = sin(x)/ cos(x), and the reciprocal rule, show
that
d
sec(x) = sec(x) tan(x).
dx
d
d
1
1
d
1
sec(x) =
=
·
cos(x) =
· ( sin(x))
dx
dx cos(x)
(cos(x))2 dx
(cos(x))2
sin(x)
1
sin(x)
=
=
·
= sec(x) tan(x).
2
(cos(x))
cos(x) cos(x)
1
Math 1310: CSM
MORE sample exam 2 problems
SOLUTIONS
dy
4. For each of the following pairs of functions y and u, find
. Make sure to express your final
dx
answer in terms of the variable x only.
(a) y =
1
, u=x+2
u
dy
1
= 2,
du
u
(b) y =
p
3
u, u = x2
du
= 1,
dx
dy
dy du
1
1
=
·
= 2 ·1=
.
dx
du dx
u
(x + 2)2
3
dy
1
= 2/3 ,
du
3u
du
= 2x,
dx
dy
dy du
1
2x
2x
=
·
= 2/3 · 2x = 2/3 =
.
dx
du dx
3u
3u
3(x2 3)2/3
(c) y = 2u , u = x + cos(x)
dy
= k2 ·2u ,
du
du
= 1 sin(x),
dx
(d) y = 5 + tan(u), u = x4 + 3x
dy
= sec2 (u),
du
du
= 4x3 +3,
dx
dy
dy du
=
·
= k2 ·2u ·(1 sin(x)) = k2 ·2x+cos(x) ·(1 sin(x)).
dx
du dx
1
dy
dy du
=
·
= sec2 (u)·(4x3 +3) = (4x3 +3) sec2 (x4 +3x 1).
dx
du dx
2
Math 1310: CSM
MORE sample exam 2 problems
5. Let
f (x, y) = x2
(a) Find fx (x, y). =
3
(b) Find fy (x, y). =
x2
2x
p
y
2/3
p
y
2/3
p
1.2
1/3
⇡ (3)2
p
1
.
.
(c) Use the full microscope equation to estimate
p
(3.1)2
1.2
(3.1)2
1/3
.
1
p
6 y x2
p
y
SOLUTIONS
1/3
+
1/3
2(3)
p
3 32
1
.
2/3
· (0.1)
p
6 1 32
1
p
1
2/3
· (0.2)
1
1
· 0.1
· 0.2 = 2.04167.
2
24
(Compare with the ”true” value of 2.04199.)
=2+
6. Fill in the blanks (there are six of them) on the picture below, by inserting exactly one of the
terms (i)–(viii) in each blank.
(i)
x
(ii)
(iii) slope (iv) one (v) secant
y
(vi) tangent
(vii) two
(viii) f 0 (x)
Note: there are more terms than there are blanks, so some terms will not be used.
tangent line: slope f’(x)
P
y
secant
line: slope ∆y/∆x
∆y
}
}
one sided
∆x
y=f(x)
x
x
x+∆x
The above point P has coordinates (x+∆x,f(x)+f’(x) ∆x )
3
Math 1310: CSM
MORE sample exam 2 problems
SOLUTIONS
7. Consider the function f (x) sketched below.
y
1.0
0.5
!1.0
!0.5
0.5
1.0
0.5
1.0
x
!0.5
!1.0
On the axes below, sketch the graph of f 0 (x).
y
15
10
5
!1.0
!0.5
!5
!10
4
x
Math 1310: CSM
MORE sample exam 2 problems
SOLUTIONS
8. Consider the function f (x) sketched below.
y
2.0
1.5
1.0
0.5
!2
!1
1
2
1
2
x
!0.5
!1.0
On the axes below, sketch the graph of f 0 (x).
3
y
2
1
-2
-1
-1
-2
-3
5
x
Math 1310: CSM
MORE sample exam 2 problems
SOLUTIONS
9. On the graph below, draw a picture indicating what approximation to f (a + 2 x) you would
get using Euler’s method (starting at x = a and using stepsize x).
f (x)
a
a+Δx
6
a+2Δx
x