THE CHINESE UNIVERSITY OF HONG KONG
Department of Mathematics
MATH1510 Calculus for Engineers (Fall 2016)
Suggested solutions of coursework 4 (Take home)
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Total
/100 pts
2
1. The left derivative of a function f (x) at x = c is by definition:
Lf 0 (c) = lim−
h→0
f (c + h) − f (c)
;
h
and the right derivative at x = c is:
Rf 0 (c) = lim+
h→0
f (c + h) − f (c)
.
h
The function f (x) is differentiable at x = c if and only if its left and right derivatives
are finite and equal to each other. If Lf 0 (c) = Rf 0 (c) and is finite, then f 0 (c) exists
and is equal to their common value.
Let
(
(t − 1)2 − 1, if − 1 ≤ t < 0;
k(t) =
tan t,
if 0 ≤ t ≤ π/4.
(a) Find Lk 0 (0).
(b) Find Rk 0 (0).
(c) Is k(t) differentiable at t = 0?
3
(a)
k(0 + h) − k(0)
h→0
h
((h − 1)2 − 1) − tan(0)
= lim−
h→0
h
2
h − 2h
= lim−
h→0
h
= lim− (h − 2)
Lk 0 (0) =
lim−
h→0
= −2
(b)
Rk 0 (0) =
=
=
=
=
tan(0 + h) − tan(0)
h→0
h
tan(h)
lim
h→0+
h
sin(h)
1
lim
·
h→0−
h
cos h
sin(h)
1
· lim−
lim−
h→0
h→0 cos h
h
1
lim+
(c) k(t) is not differentiable at t = 0, since the right and left derivatives at
t = 0 are not equal to each other.
4
2. Let us study the derivative as a function. The function f 0 (x) is still defined as a limit,
but the fixed number a is replaced by the variable x :
f (x + h) − f (x)
.
h→0
h
f 0 (x) = lim
(1)
If y = f (x), we also write y 0 or y 0 (x) for f 0 (x). The domain of f 0 (x) consists of all
values of x in the domain of f (x) for which the limit in Equation (1) exists. We say
that f (x) is differentiable on (a, b) if f 0 (x) exists for all x in (a, b). When f 0 (x) exists
for all x in the interval or intervals on which f (x) is defined, we say simply that f (x) is
differentiable.
Using Equation (1), determine the domain of f 0 , then give a formula describing
f 0 (x) where
√
f (x) = x2 − 1 with domain Df = (−∞, −1] ∪ [1, ∞).
5
For any x ∈ Df \ {−1, 1} ,
f (x + h) − f (x)
f 0 (x) = lim
h→0
h
p
√
2
(x + h) − 1 − x2 − 1
= lim
h→0
h
p
p
√
√
2
(x + h) − 1 − x2 − 1 (x + h)2 − 1 + x2 − 1
p
= lim
√
h→0
h
(x + h)2 − 1 + x2 − 1
p
√
2
2
(x + h)2 − 1 − x2 − 1
= lim p
√
h→0 h( (x + h)2 − 1 +
x2 − 1)
√2
(x + h)2 − 1 − x2 + 1
= lim p
because t = t
√
h→0 h( (x + h)2 − 1 +
x2 − 1)
2xh + h2
= lim p
√
h→0 h( (x + h)2 − 1 +
x2 − 1)
2x + h
= lim p
√
h→0
(x + h)2 − 1 + x2 − 1
x
=p
(x)2 − 1
At x = −1, we need to consider the left-derivative :
Lf 0 (−1) = lim−
h→0
f (−1 + h) − f (−1)
h
By the same computations with x = −1, we have
2(−1) + h
−2 + h
p
Lf 0 (−1) = lim− p
= lim− √
2
2
h→0
h2 − 2h
(−1 + h) − 1 + (−1) − 1 h→0
which does not exist. Similarly, at x = 1, we have
2(1) + h
2+h
p
Rf 0 (1) = lim+ p
= lim+ √
h→0
h2 + 2h
(1 + h)2 − 1 + (1)2 − 1 h→0
which also does not exist.
Hence, the domain of f 0 is (−∞, −1) ∪ (1, ∞).
6
3. In the table below, f (x) and g(x) are differentiable at x:
Derivative of a constant:
d
(c) = 0
dx
(c, a constant).
d n
(x ) = nxn−1 .
dx
d
d
Derivative of a constant multiple of a function:
[cf (x)] = c [f (x)]
dx
dx
(c, a constant).
d
d
d
Sum rule/Difference rule:
[f (x) ± g(x)] =
[f (x)] ± [g(x)].
dx
dx
dx
d
d
d
Product rule:
[f (x)g(x)] = f (x) [g(x)] + g(x) [f (x)].
dx
dx
dx
d
d
g(x) [f (x)] − f (x) [g(x)]
d f (x)
dx
dx
=
(g(x) 6= 0).
Quotient rule:
2
dx g(x)
[g(x)]
d
df (u) dg(x)
Chain rule:
f (g(x)) =
·
.
dx
du u=g(x) dx
Power rule: If n is any real number, then
Let a be a constant with a > 0, a 6= 1.
(if a = e)
(if a = e)
d n
x = nxn−1
dx
d n
a = (ln a)ax
dx
d x
e = ex
dx
d
1
1
loga x =
dx
ln a x
1
d
ln x =
dx
x
d
sin x = cos x
dx
d
cos x = − sin x
dx
d
tan x = sec2 x
dx
d
cot x = − csc2 x
dx
d
sec x = sec x tan x
dx
d
csc x = − csc x cot x
dx
d
d
1
sin−1 x =
arcsin x = √
dx
dx
1 − x2
d
d
1
tan−1 x =
arctan x =
dx
dx
1 + x2
d n
du
u = nun−1
dx
dx
d u
du
a = (ln a)au
dx
dx
d u
du
e = eu
dx
dx
d
1 du
loga u =
dx
(ln a)u dx
1 du
d
ln(u) =
dx
u dx
d
du
sin u = cos u
dx
dx
d
du
cos u = − sin u
dx
dx
d
du
tan u = sec2 u
dx
dx
d
du
cot u = − csc2 u
dx
dx
d
du
sec u = sec u tan u
dx
dx
d
du
csc u = − csc u cot u
dx
dx
d
d
1
du
sin−1 u =
arcsin u = √
2
dx
dx
1 − u dx
d
d
1 du
tan−1 u =
arctan u =
dx
dx
1 + u2 dx
7
Answer the following questions :
(a) Differentiate the following functions with respect to x:
i. y = (ln x) (tan x) + (sin x) (cos x);
3x2 + 3x − 9
ii. y =
;
1 + ex
iii. y = (3x − 2)3 (3 − x2 )2 ;
iv. y = sec(xe ) + tan(ex );
sin−1 x
+ sec x csc x;
v. y =
e2x
vi. y = log2 (ln x + sin(2x )).
dy
, where the variables x, y satisfy:
(b) By implicit differentiation, find
dx
xy = ex sin y.
You may express your answer in terms of x and y.
(a)(i).
y0 =
1
· tan(x) + ln x · sec2 x + (cos x)(cos x) + (sin x)(− sin x)
x
1
= · tan(x) + ln x · sec2 x + cos2 (x) − sin2 (x)
x
(a)(ii).
(6x + 3)(1 + ex ) − (3x2 + 3x − 9)ex
(1 + ex )2
3(ex (−x2 + x + 4) + 2x + 1)
=
(1 + ex )2
y0 =
(a)(iii).
y 0 = 3(3x − 2)2 · 3 · (3 − x2 )2 + (3x − 2)3 · 2(3 − x2 ) · (−2x)
= 9(3x − 2)2 (3 − x2 )2 − 4x(3x − 2)3 (3 − x2 )
= (3x − 2)2 (3 − x2 )(27 + 8x − 21x2 )
8
(a)(iv).
Since
(sec x)0 = sec x tan x, (tan x)0 = sec2 x
we have
y 0 = exe−1 sec(xe ) tan(xe ) + ex sec2 (ex )
(a)(v).
We write y as following:
y = e−2x sin−1 x + sec x csc x
since
(sin−1 x)0 = √
1
, (csc x)0 = − csc x cot x
1 − x2
we have
e−2x
y0 = √
− 2e−2x sin−1 x + (sec x tan x) csc x − sec x(csc x cot x)
1 − x2
(a)(vi).
0
y =
d
dx
(ln x + sin(2x ))
=
ln x + sin(2x ) ln 2
1
x
+ cos(2x )2x ln 2
ln x + sin(2x ) ln 2
(b).
Differentiate both sides with respect to x,
d x
d
(xy) =
(e sin y)
dx
dx
y + xy 0 = ex sin y + ex (cos y)y 0
y0 =
ex sin y − y
x − ex cos y
9
4. (Optional) Answer the following questions:
(a) The hyperbolic sine function, the hyperbolic cosine function and the hyperbolic tangent function are defined as the following combinations of exponential
functions:
sinh x =
ex − e−x
,
2
cosh x =
ex + e−x
,
2
tanh x =
Show that
d
sinh x = cosh x;
i.
dx
d
ii.
cosh x = sinh x;
dx
d
iii.
tanh x = sech2 x;
dx
d
iv.
coth x = −csch2 x;
dx
d
v.
sech x = −sech x tanh x;
dx
d
vi.
csch x = −csch x coth x.
dx
(b) Differentiate
√
i. y = sinh x + 1;
ii. y = coth x3 ;
iii. y = coth (tanh 3x) + sech (coth 4x).
(a)
i.
d ex − e−x
ex + e−x
d
sinh x =
=
= cosh x
dx
dx
2
2
ii.
d
d ex + e−x
ex − e−x
cosh x =
=
= sinh x
dx
dx
2
2
iii.
d
d sinh x
tanh x =
dx
dx cosh x
d
d
cosh x dx
sinh x − sinh x dx
cosh x
=
2
cosh x
cosh2 x − sinh2 x
=
cosh2 x
= sech2 x (since cosh2 x − sinh2 x = 1)
ex − e−x
.
ex + e−x
10
iv.
d cosh x
d
coth x =
dx
dx sinh x
d
d
cosh x − cosh x dx
sinh x
sinh x dx
=
2
sinh x
2
sinh x − cosh2 x
=
sinh2 x
= − csch2 x
v.
d
cosh x
d
d
1
sinh x
sech x =
(cosh x)−1 = − dx 2
·
= − sech x tanh x
=−
dx
dx
cosh x cosh x
cosh x
vi.
d
sinh x
d
d
cosh x
1
csch x =
(sinh x)−1 = − dx 2
=−
·
= − csch x coth x
dx
dx
sinh x sinh x
sinh x
(b)
i.
√
√
d√
1
x+1= √
cosh x + 1
y 0 = cosh x + 1
dx
2 x+1
ii.
y 0 = − csch2 x3
d 3
x = −3x2 csch2 x3
dx
iii.
d
coth 4x
dx
= −3 sech2 (3x) csch2 (tanh 3x) + 4 sech(coth 4x) tanh(coth 4x) csch2 4x
y 0 = − csch2 (tanh 3x) · sech2 (3x) · 3 − sech(coth 4x) tanh(coth 4x)