MATH6501 - Autumn 2015
Solutions to Problem Sheet 1
1. (a) First, let y = ln(sin (x2016 )). Put
f (u) = ln u,
u(x) = sin (x2016 ),
then by the Chain Rule,
dy
df du
1 d
=
=
sin (x2016 ) .
dx
du dx
u dx
Next, substitute for u in terms of x. This gives
dy
1
d
=
sin (x2016 ) .
2016
dx
sin (x
) dx
Now apply the Chain Rule again to compute
d
sin (x2016 ) ,
dx
this time with
f (u) = sin u,
u(x) = x2016 .
Then
d
sin (x2016 ) = (cos u)(2016 x2015 ).
dx
= 2016 x2015 cos (x2016 ).
Finally, putting everything together, one deduces
that
dy
cos (x2016 )
= 2016 x2015
dx
sin (x2016 )
2015
= 2016 x
cot (x2016 ).
MATH6501 - Autumn 2015
(b) The easiest route for y = (cos x)x would be to start
with logarithmic differentiation, so take logs on both
sides of this equation to get
ln y = x ln(cos x).
Differentiating both sides w.r.t. x gives the following. . .
d
1 dy
=
(x ln(cos x))
y dx
dx
d
= 1 · ln(cos x) + x ·
(ln(cos x))
dx
−sin x
= ln(cos x) + x
cos x
where the Product and then the Chain Rule have
been applied. Therefore
1 dy
= ln(cos x) − x tan x,
y dx
and so
dy
= y [ln(cos x) − x tan x]
dx
= (cos x)x [ln(cos x) − x tan x] .
(c) First of all, note that cos3 (exp x) is the same as
cos3 (ex ). Then to compute the derivative of y =
cos3 (ex ) by direct use of the Chain Rule, start by
taking
f (u) = u3 ,
u(x) = cos (ex ),
MATH6501 - Autumn 2015
then
d
dy
= 3u2
(cos (ex ))
dx
dx
d
= 3cos2 (ex ) (cos (ex )) .
dx
Next, use the Chain Rule a second time, this time
with
f (u) = cos u,
u(x) = ex
hence
d
d
(cos (ex )) = −sin u (ex ) = −sin (ex )ex .
dx
dx
Putting everything together, one has that
dy
= −3ex cos2 (ex )sin (ex ).
dx
(d) The easiest method for computing the derivative of
sin x
y= 2
is by using the Quotient Rule with
x + sin x
u = sin x,
du
= cos x,
dx
This gives
v = x2 + sin x,
dv
= 2x + cos x.
dx
cos
x)
dy
(x2 + sin
x) · cos x − sin x · (2x + =
2
2
dx
(x + sin x)
x2 cos x − 2x sin x
=
(x2 + sin x)2
x(xcos x − 2 sin x)
=
.
(x2 + sin x)2
MATH6501 - Autumn 2015
(e) Use the Product Rule with
u = e7x ,
v = 2 sin (10x) + cos (10x),
du
dv
= 7e7x ,
= 20 cos (10x) − 10 sin (10x).
dx
dx
Thus
dy
= e7x [20 cos (10x) − 10 sin (10x)]
dx
+ 7e7x [2 sin (10x) + cos (10x)]
= e7x [(14 − 10) sin (10x) + (20 + 7) cos (10x)] .
So
dy
= e7x [4 sin (10x) + 27 cos (10x)] .
dx
(f) Start with the Chain Rule where
f (u) = sin u,
u(x) = cos (ex ln x) ,
then
dy
du
= cos u
dx
dx
d
(cos (ex ln x)) .
dx
Next, use the Chain Rule again, this time with
= cos(cos (ex ln x))
f (u) = cos u, u(x) = ex ln x,
to yield
du
d
(cos (ex ln x)) = −sin u
dx
dx
d x
(e ln x)
dx
1
= −sin (ex ln x) ex + ex ln x ,
x
= −sin (ex ln x)
MATH6501 - Autumn 2015
where the Product Rule has been used for the last
step.
Hence
dy
x
x
x1
x
= −cos(cos (e ln x))sin (e ln x) e + e ln x
dx
x
ex
= − (1 + x ln x)sin (ex ln x)cos(cos (ex ln x)).
x
2. If w(x) =
1
v(x) ,
then the Product Rule states that:
f 0 (x) = u0 (x)w(x) + u(x)w0 (x)
= u0 (x) ·
1
d
+ u(x) ·
v(x)
dx
(1)
1
v(x)
.
(2)
But by the Chain Rule,
d
1
−1
−v 0 (x)
0
=
·
v
(x)
=
.
dx v(x)
[v(x)2 ]
[v(x)2 ]
Thus
0
−v (x)
u0 (x)
+ u(x) ·
f (x) =
v(x)
[v(x)2 ]
u0 (x) u(x)v 0 (x)
=
−
v(x)
[v(x)2 ]
0
v(x)u (x) u(x)v 0 (x)
=
−
.
[v(x)2 ]
[v(x)2 ]
0
Therefore
f 0 (x) =
v(x)u0 (x) − u(x)v 0 (x)
,
[v(x)]2
which happens to be the Quotient Rule!
(3)
(4)
(5)
MATH6501 - Autumn 2015
3. (a) The Quotient Rule will be used here. . .
d
d sin x
(tan x) =
dx
dx cos x
d
d
cos x dx
(sin x) − sin x dx
(cos x)
2
cos x
cos x × cos x − sin x × (− sin x)
=
cos2 x
2
2
cos x + sin x
=
cos2 x
1
=
[∵ cos2 x + sin2 x ≡ 1]
cos2 x
1
]
= sec2 x,
[∵ sec x ≡
cos x
=
which is the desired result.
(b) Put y = tan−1 x. Then
tan y = x.
(6)
Next, both sides can be differentiated implicitly
(with the aid of the result from part a). The result
is that. . .
1 = sec2 y
dy
dx
dy
1
=
.
dx
sec2 y
But sec2 y ≡ 1 + tan2 y, hence
dy
1
=
.
dx
1 + tan2 y
MATH6501 - Autumn 2015
Finally, one can substitute in Equation (6) to rewrite
the derivative in terms of x. This gives the expected
result of
dy
1
=
.
dx
1 + x2
4. Implicitly differentiate both sides of the equation w.r.t.
x, i.e. compute
d 3
d
d
d
d
d
(x )− (3xy 2 )+ (2y)− (ex )+ (1) =
(cos y).
dx
dx
dx
dx
dx
dx
This reduces to
dy
dy
dy
3x2 − 3y 2 − 6xy
+2
− ex + 0 = − sin y .
dx
dx
dx
Next, the reader is advised to collect all the
the right hand side as follows:
3x2 − 3y 2 − ex = (6xy − 2 − sin y)
dy
dx
terms on
dy
,
dx
which rearranges to
dy
3x2 − 3y 2 − ex
=
.
dx
6xy − 2 − sin y
5. (a) Look closely at the function
R(p) = p · f (p).
This is a product of two functions, thus the Product
Rule comes in handy here. One finds that
R0 (p) = 1 · f (p) + p · f 0 (p)
= f (p) + p · f 0 (p).
MATH6501 - Autumn 2015
Next, let p = 120.
R0 (120) = f (120) + 120f 0 (120).
Thankfully, we are given that f (120) = 9000 and
f 0 (120) = −60. If we substitute in these numbers,
one sees that:
R0 (120) = 9000 − (120 × 60)
= 9000 − 7200
= 1800.
(b) Because R0 (120) > 0, a small increase in price will
result in an increase of the manufacturer’s revenue.