Tutorial 6: SOLUTIONS
Derivatives 2
1. For the following functions, we must compute dy/dx:
(a) y = sin(1/x2 ).
Applying the chain rule gives
dy
−2 cos(1/x2 )
.
= cos(1/x2 ) · (−2x−3 ) =
dx
x3
2
1+csc(x )
(b) y = 1−cot(x
2) .
We may re-express the function by noting that csc x = 1/ sin x and cot x =
cos x/ sin x, and then multiplying above and below by sin x gives
y(x) =
sin(x2 ) + 1
.
sin(x2 ) − cos(x2 )
Differentiating using both the quotient and the chain rule
cos(x2 ) · (2x)[sin(x2 ) − cos(x2 )] − (cos(x2 ) · (2x) + sin(x2 ) · (2x))[sin(x2 ) + 1]
dy
=
dx
(sin(x2 ) − cos(x2 ))2
−2x(1 + cos(x2 ) + sin(x2 ))
=
(sin(x2 ) − cos(x2 ))2
−2x(1 + cos(x2 ) + sin(x2 ))
=
1 − 2 sin(x2 ) cos(x2 )
where in the last line we have multiplied out the denominator explicitly and
used the fact that cos2 (x2 ) + sin2 (x2 ) = 1.
(c) y = cos(cos(x)).
Here we will use the formal version of the chain rule. Defining u = cos x, then
y = cos u. Differentiating with respect to x gives
dy
dy du
d(cos u) d(cos x)
=
=
= (− sin u)(− sin x) = sin(cos x) sin x.
dx
du dx
du
dx
2. We want to use the fact that
d
(|x|) =
dx
(
1
−1
as well as the chain rule to find
d
(| sin x|)
dx
1
x>0
x<0
for non-zero x in the interval (−π, π). Again, we adopt the more formal approach
to the chain rule by defining u = sin x. Then we have
d(|u|)
d(|u|) du
d(|u|)
=
=
cos x.
dx
du dx
du
Since d(|u|)/du = 1 for u > 0 and d(|u|)/du = −1 for u < 0, we obtain
(
cos x
u>0
d
(| sin x|) =
dx
− cos x
u < 0.
But on the interval (−π, π)
u = sin x > 0
⇐⇒
0<x<π
u = sin x < 0
⇐⇒
−π < x < 0.
Hence, we have
d
(| sin x|) =
dx
(
cos x
− cos x
0<x<π
−π < x < 0.
3. We are asked to implicitly differentiate the following in order to find d2 y/dx2 :
(a) 2x2 − 3y 2 = 4. Differentiating once and using the chain rule gives
4x − 6y
dy
=0
dx
=⇒
dy
2x
=
.
dx
3y
Differentiating this again gives
d2 y
2(3y) − 3(dy/dx)(2x)
6y − 6x(2x/3y)
6y 2 − 4x2
=
=
=
.
dx2
9y 2
9y 2
9y 3
Finally, we can simplify the top line considerably by multiplying the original
equation by -2 which gives
6y 2 − 4x2 = −8
and subbing into our expression for the second derivative yields
d2 y
8
= − 2.
2
dx
9y
(b) xy + y 2 = 2. Differentiating once, using the product and chain rule gives
y+x
dy
dy
+ 2y
=0
dx
dx
2
=⇒
dy
y
=−
.
dx
x + 2y
Differentiating again gives
(dy/dx)(x + 2y) − (1 + 2dy/dx)(y)
d2 y
=−
2
dx
(x + 2y)2
y
y
(− x+2y )(x + 2y) − (1 + 2(− x+2y
))y
=−
(x + 2y)2
−y(x + 2y) − yx
=−
(x + 2y)3
2(xy + y 2 )
=
.
(x + 2y)3
Finally, we simplify the top line using the original equation yielding
4
d2 y
=
.
2
dx
(x + 2y)3
(c) x cos y = y. Differentiating once, using the product and chain rule, we get
cos y − x sin y
dy
dy
=
dx
dx
=⇒
dy
cos y
=
.
dx
1 + x sin y
Differentiating again gives
dy
dy
− sin y dx
(1 + x sin y) − (sin y + x cos y dx
) cos y
d2 y
=
2
2
dx
(1 + x sin y)
cos y
cos y
− sin y( 1+x sin y )(1 + x sin y) − (sin y + x cos y( 1+x
sin y )) cos y
=
2
(1 + x sin y)
− sin y cos y(1 + x sin y) − cos y sin y − x cos y
=
(1 + x sin y)3
2 sin y cos y + x cos y(1 + sin2 y)
=−
(1 + x sin y)3
In this case there is no real obvious simplification, we could for example
replace the x cos y by y in the numerator since they are equal according to
the original equation, which gives
2 sin y cos y + y(1 + sin2 y)
d2 y
=−
.
2
dx
(1 + x sin y)3
4. A 17-ft ladder is leaning against a wall. If the bottom of the ladder is pulled along
the ground away from the wall at a constant rate of 5 ft/s, how fast will the top of
the ladder be moving down the wall when it is 8 ft above the ground. The diagram
is given below. We want to know dh/dt when h = 8 given that dl/dt = 5 for all
time. The quantities are related by Pythagoras theorem:
h2 + l2 = 172 .
3
17
h
l
Differentiating with respect to t gives
2h
dh
dl
+ 2l = 0.
dt
dt
Since dl/dt = 5, we get
5l
dh
=−
dt
h
where the minus sign is because the height is decreasing with increasing time.
When h = 8, we can work out l using Pythagoras again
l2 = 172 − 82
Hence
dh =⇒
l = 15.
5(15)
= −9.375.
dt h=8
8
So when the height is 8f t, the ladder is dropping at a rate of 9.375f t/s.
=−
5. The side of a cube is measured with a possible percentage error of ±2%. We are
asked to use differentials to estimate the percentage error in the volume. The
percentage error of a differentiable function f is df /f . We are told that
dx
= ±0.02.
x
We want to estimate the percentage error in the volume, which is given by dV /V ,
where
V = x3 , =⇒
dV = 3x2 dx.
Therefore
dV
dx
=3
= 3(±0.02) = ±0.06,
V
x
i.e., the percentage error in the volume is ±6%.
4