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Test 
Winter 
Calculus I (-nya/)
Question . — Compute each derivative.
dy
x 5
a. Compute
, where y = x5 − 5x + − + 55 − xx + πe .
dx
5 x
q tan q
dp
, where p = (q3 − q2 + 7)2/3 sin e
.
b. Compute
dq
q
sin5 (y 2 − 3) 3y 2 + 4
dw
, where w =
.
c. Compute
2
dy
y 4 esec (y) (2y + 1)2/3
Solution. — a. If −1 < x < 1 and x , 0, then 1 + x < ex and (replacing x by −x) 1 − x < e−x , or
ex < 1/(1 − x). Subtracting x yields x < ex − 1 < x/(1 − x), and then dividing by x and replacing x
by x0 − x yields
0
ex −x − 1
1
1<
<
if
x < x0 < x + 1
x0 − x
1 − x0 + x
and
0
1
ex −x − 1
<
<1
if
x − 1 < x0 < x.
1 − x0 + x
x0 − x
Since 1/(1 − x0 + x) → 1 as x0 → x, it follows that
0
0
d x
ex −x − 1
ex − ex
= ex · lim
= ex · 1 = ex .
e = lim 0
dx
x0 →x x0 − x
x0 →x x − x
dr
2
d. Compute
, where ear +b sin(t 3 ) − cot(r 2 t 4 ) + r 3 = t 7 + 9.
dt
dy
, where y = cos(xsin x ).
dx
dy
1
5
Solution. — a.
= 5x4 − 5x log 5 + + 2 − xx (1 + log x)
dx
5 x
b. The first limit is zero by elementary properties of the logarithm. The second limit is
e. Compute
dp 2(3q2 − 2q) sin(eq tan q )
b.
=
+ (q3 − q2 + 7)2/3 cos(eq tan q )eq tan q (tan q + q(1 + tan2 q))
dq
3(q3 − q2 + 7)1/3
q
)
(
5 2
2
3y
4
dw sin (y − 3) 3y + 4 10y cos(y 2 − 3)
4
2 (y) tan(y) −
=
−
2
sec
c.
·
+
−
2
dy
3(2y + 1)
sin(y 2 − 3)
3y 2 + 4 y
y 4 esec (y) (2y + 1)2/3
2
3t 2 ear +b cos(t 3 ) + 4r 2 t 3 csc2 (r 2 t 4 ) − 7t 6
dr
d.
=−
2
dt
2arear +b sin(t 3 ) + 2rt 4 csc2 (r 2 t 4 ) + 3r 2
e.
dy
= − sin(xsin x )(xsin x ) · (cos x log x + (sin x)/x)
dx
dy
x
, where y =
.
dx
1 + x2
x
b. Find an equation of the line tangent to the curve defined by y =
, where x = 2.
1 + x2
lim
Solution. — a. If y =
y0 − y =
x
and y 0 =
, then
1 + x2
1 + x0 2
x0 (1 + x2 ) − x(1 + x0 2 )
(x0 − x)(1 − xx0 )
x
−
=
=
.
1 + x0 2 1 + x2
(1 + x0 2 )(1 + x2 )
(1 + x0 2 )(1 + x2 )
x0
Therefore,
y0 − y
dy
1 − xx0
1 − x2
= lim 0
= lim
=
,
2
0
0
0
2
dx x →x x − x x →x (1 + x )(1 + x ) (1 + x2 )2
by independence and direct substitution.
dy
3 , so the tangent line is defined by 3x + 25y = 16.
b. If x = 2 then y = 25 and dx = − 25
Question . — a. Given 1 + x < ex for x , 0, use the definition of the derivative to prove that
d x
e = ex .
dx
b. Compute each of the following limits:
n√
o
ln(x + 2)
2x − 3 x/2
lim
x (ln x)23 ;
lim
;
lim
.
√
x→∞ 2x + 1
x→−1 2 − 3x + 7
x→0+
= −1
and
lim (1 + t)1/t = e,
t→0
it follows that
! 2x+1 · −2x
2x+1
−4
−4
= e−1 .
x→∞
2x + 1
After rationalizing the denominator, the third limit becomes
√
(
)
log(y)
log(x + 2) 2 + 3x + 7
·
= − 43 ,
= 1.
since
lim
lim
x+1
−3
x→−1
y→1 y − 1
lim 1 +
Question . — The curve defined by x3 y 2 − x2 y 3 = k passes through the point (2, 1).
dy
a. What is the value of k?
b. Find dx and simplify your answer.
c. Find all points on the curve where the tangent line has slope 1.
Solution. — a. Since 23 · 12 − 22 · 13 = 4, it follows that k = 4.
b. Implicit differentiation gives
dy
3x2 y 2 − 2xy 3 2y 2 − 3xy y(2y − 3x)
=
=
=− 3
.
dx
2x y − 3x2 y 2 2x2 − 3xy x(2x − 3y)
Question . — a. Use the definition of the derivative to compute
x0
−2x
x→∞ 2x + 1
c. The tangent line has slope 1 where 2y 2 − 3xy = 2x2 − 3xy (and xy , 0), or x2 = y 2 , which is
where y = x or y = −x. There are no points on the curve where x = y (since zero is not equal to 4).
Replacing y by −x in the equation of the curve yields
x3 (−x)2 − x2 (−x)3 = 4,
or equivalently,
2x5 = 4,
√
√
5
5
so x = 2, and y = − 2. Therefore, the tangent line to the curve has slope 1 at the one point
√
√
( 5 2, − 5 2).
Question . — The line 2x + 3y = 5 is tangent
to the graph of f where x = 2, and f 00 (2) = −6. The
function g is defined by g(x) = sin πf (x) .
√
a. Write an equation of the line tangent to the graph of y = g( x) at the point where x = 4.
b. Evaluate g 00 (2).
Solution. — a. Since 2·2+3· 13 = 5 and the slope of the given line is − 23 , f (2) = 13 and f 0 (2) = − 23 .
√
Also, g(2) = sin 31 π = 12 3, and g 0 (2) = π cos 31 π · − 32 = − 13 π. Therefore,
√ √ √
dy g 0 ( x) g 0 (2)
1 π.
√
y = g( x) = g(2) = 21 3,
and
=
=
= − 12
dx
4
2 x
x=4
x=4
x=4
x=4
√
1 π(x − 4).
Thus, the tangent line to the curve is defined by y = 12 3 − 12
Test 
b. Since g 0 (x) = π cos πf (x) f 0 (x), it follows that
2
√ 2
g 00 (2) = −π2 sin πf (2) f 0 (2) + π cos πf (2) f 00 (2) = −π2 · 12 3 · − 32 + π · 21 · (−6)
√
= − 29 π2 3 − 3π.
Question . — For each equation, find all points on its graph where the tangent line is horizontal.
a. y = x
(ln x)/x
b. y = (sin(x) + 1) cos(x)
Solution. — a. Logarithmic differentiation gives
(log x)/x d (log x)2
(log x)/x
(log x)/x (log x)(2 − log x)
dy
2 log x (log x)2
=x
=x
·
−
=x
·
,
2
2
dx
dx
x
x
x
x2
which is zero if, and only if, either log x = 0, i.e., x = 1, or else log x = 2, i.e., x = e2 . If x = 1 then
2
2
y = 1, and if x = e2 then y = (e2 )2/e = e4/e . Therefore, the tangent line to the curve is horizontal
2
at the points (1, 1) and (e2 , e4/e ).
b. The product rule and the Pythagorean identity gives
dy
= −(sin(x) + 1) sin(x) + cos2 (x) = 1 − sin(x) − 2 sin2 (x) = (1 + sin(x))(1 − 2 sin(x)),
dx
which is equal to zero if, and only if, either sin(x) = −1 or else sin(x) = 12 . The solutions of
sin(x) = −1 are 32 π + 2kπ, where k is an integer, and the corresponding values of y are all zero.
The solutions of sin(x) = 12 are 16 π + 2kπ and 56 π + 2kπ, where k is an integer. If x = 16 π + 2kπ
√
√
then y = 34 3, and if x = 56 π + 2kπ then y = − 34 3. Therefore, the tangent line to the curve
√ √ is horizontal at the points 32 π + 2kπ, 0 , 16 π + 2kπ, 34 3 and 65 π + 2kπ, − 34 3 , where k is an
integer.
Question . — The curve defined by (x − 2)y = x has two tangent lines which contain the point
(−1, 3). Write an equation of each of these tangent lines.
Solution. — If
dy
x
2
,
then
=−
,
x−2
dx
(x − 2)2
so the tangent line to the curve at a point (x, y) contains the point (−1, 3) if, and only if,
y=
x
2
−3 = −
(x + 1),
x−2
(x − 2)2
or equivalently
x(x − 2) − 3(x − 2)2 = −2(x + 1).
Collecting terms and factorizing gives 0 = 2x2 − 12x + 10 = 2(x − 1)(x − 5), whose solutions are 1
dy
and 5. If x = 1 then y = −1 and dx = −2, so the tangent line to the curve at (1, −1) is defined by
dy
y = −1 − 2(x − 1), or 2x + y = 1. If x = 5 then y = 53 and dx = − 29 , so the tangent line to the curve
at 5, 53 is defined by y = 53 − 92 (x − 5), or 2x + 9y = 25. Therefore, the tangent lines to the curve
which contain the point (−1, 3) are defined by 2x + y = 1 and 2x + 9y = 25.
Question . — An object moves along a vertical line according to the equation
y = 2t 3 − 9t 2 + 12t − 1,
where
t > 0.
Units are irrelevant. If you must have units, then y is measured in furlongs and t is measured in
fortnights. But it is not necessary to write units.
a.
b.
c.
d.
Winter 
Calculus I (-nya/)
Find the velocity and acceleration functions of the object.
Find the time intervals during which the object is rising.
Find the velocity of the object when its acceleration is zero.
Find the distance travelled by the object while 0 6 t 6 3.
Solution. — a. The velocity and acceleration functions of the object are given by
dy
d2 y
= 6t 2 − 18t + 12 = 6(t − 1)(t − 2),
and
a = 2 = 6(2t − 3).
dt
dt
b. The object is rising when its velocity is positive, which is when t − 1 and t − 2 are either
both positive or both negative. So the object is moving upward if 0 6 t < 1 or else t > 2. c. The object’s acceleration is zero when t = 23 , at which time its velocity is 6 12 − 21 = − 32 .
d. While 0 < t < 1 the object rises from y0 = −1 to y1 = 4, while 1 < t < 2 the object falls from
y1 = 4 to y2 = 3, and while 2 < t 6 3 the object rises from y2 = 3 to y3 = 8. Therefore, during the
first three fortnights the object travels 5 + 1 + 5 = 11 furlongs.
v=
Question . — Find the 87th derivative,
d87y
, where y = 3x23 + sin(3x) − xe2x .
dx87
Solution. — The 87th derivative of 3x23 is zero, since it is a polynomial of degree smaller than
87, and the 87th derivative of sin(3x) is −387 cos(3x) by the chain rule, since 87 is one less than a
multiple of 4. Computing the first few derivatives of z = xe2x ,
dz
= e2x (2x + 1),
dx
reveals the pattern
d2 z
= 2e2x (2x + 2)
dx2
and
d3 z
= 22 e2x (2x + 3),
dx3
dn z
= 2n−1 e2x (2x + n),
dxn
which is verified by the computation
o
d n n−1 2x
2
e (2x + n) = 2n−1 e2x 2(2x + n) + 2 = 2n e2x (2x + n + 1).
dx
Therefore,
d87y
= −387 cos(3x) − 286 e2x (2x + 87).
dx87
Question . — Use the limit definition of the derivative to prove the product rule for differentiation.
Solution. — Suppose that u and v are differentiable functions of x, and let y = uv. Then
0
dy
y0 − y
u 0 v 0 − uv 0 + uv 0 − uv
u −u 0
v0 − v
= lim 0
= lim
= lim
·v +u · 0
0
0
dx x0 →x x − x x0 →x
x −x
x −x
x0 →x x − x
du
dv
=
v +u ,
dx
dx
0
0
where v → v as x → x by the continuity of differentiable functions.