MA1S11 Tutorial Sheet 6, Solutions1
17-20 November 2015
Questions
The numbers in brackets give the numbers of marks available for the question.
1. (2) Derive the quotient rule for (f /g)0 from the product rule for (f h)0 with h = 1/g,
and the chain rule for h0 .
Solution: By the product rule we first get
(f /g)0 = (f h)0 = f 0 h + f h0
(1)
To obtain h0 we use the chain rule for h = 1/g (which is the composition of the
function 1/x with g(x))
h0 = −g 0 /g 2
(2)
so that we get
(f /g)0 = (f h)0 = f 0 h + f h0 = f 0 /g + f (−g 0 /g 2 ) =
f 0g − f g0
,
g2
(3)
i.e. the quotient rule.
2. (4) Differentiate the following functions with respect to x
√
1
,
1+x
sin (cos x),
sin
1
,
1 + x2
√
sin x,
tan 2x =
sin(2x)
cos(2x)
(4)
√
Solution: The first is again an application of the chain rule, with f (x) = 1/ x and
g(x) = x + 1. Since g 0 (x) = 1 and
f 0 (x) =
d −1/2
1
x
= − x−3/2 ,
dx
2
(5)
one has,
d
1
1
√
= f 0 (g(x))g 0 (x) = f 0 (g(x)) = − (x + 1)−3/2 .
dx 1 + x
2
(6)
For the next one we use again the chain rule
d
sin (cos x) = cos(cos x)(− sin x).
dx
1
Stefan Sint, sint@maths.tcd.ie, see also http://www.maths.tcd.ie/~sint/MA1S11.html
1
(7)
In the following example we need to apply the chain rule and then the quotient and
chain rule again
d
1
d
1
1
sin
=
cos
(8)
dx
1 + x2
1 + x2 dx 1 + x2
and then the quotient for the last term with f (x) = 1 and g(x) = 1 + x2 :
1
d
−2x
=
2
dx 1 + x
(1 + x2 )2
(9)
so that the final result is
−2x
(10)
(1 + x2 )2
√
The next example needs again the chain rule with f (u) = u and g(x) = sin(x)
1
d
sin
=
dx
1 + x2
1
cos
1 + x2
cos x
d√
sin x = √
.
dx
2 sin x
(11)
To differentiate tan 2x we need the quotient rule with f (x) = sin(2x) and g(x) =
cos(2x) and the chain rule for the derivatives f 0 and g 0 :
f 0 (x) = 2 cos(2x),
g 0 (x) = −2 sin (2x).
(12)
Then the quotient rule gives
2 cos(2x) cos(2x) + 2 sin(2x) sin(2x)
2
d sin(2x)
=
=
dx cos(2x)
cos2 (2x)
cos2 (2x)
(13)
where we have used cos2 (2x) + sin2 (2x) = 1 in the last step.
3. (2) Find dy/dx for
3yx6 − 4x2 + 6 sin y 4 = 0
(14)
Solution: Differentiate across so
d
3yx6 − 4x2 + 6 sin y 4 = 0
dx
(15)
d 6
d
d
yx − 4 x2 + 6 sin y 4 = 0
dx
dx
dx
(16)
d 6
dy
yx = 6x5 y + x6
dx
dx
(17)
d sin y 4
dy
= 4y 3 cos(y 4 )
dx
dx
(18)
giving
3
Using the product rule
and the chain rule
2
and hence
18x5 y + 3x6 y 0 − 8x + 24y 3 y 0 cos y 4 = 0
(19)
giving
y0 =
8x − 18x5 y
3x6 + 24y 3 cos y 4
(20)
Extra Questions
The questions are extra; you don’t need to do them in the tutorial class.
1. Differentiate f (x) = x−n (with n a positive integer number) in 3 different ways:
• power rule for negative integer exponent
• quotient rule for 1/f (x) and the power rule for postive exponent.
• product rule for xn f (x) and power rule for positive exponent.
Solution:
• Differentiating directly gives f 0 (x) = −nx−n−1 by the power rule with negative
integer exponent.
• Using instead the quotient rule we have
0
0 × f (x) − f 0 (x)
−f 0 (x)
1
(x) =
=
f
(f (x))2
(f (x))2
(21)
On the other hand we can directly evaluate the left hand side using the power
rule for positive integer exponents, since 1/x−n = xn ,
d n
x = nxn−1 .
dx
(22)
d n
x = −x−2n nxn−1 = −nx−n−1
dx
(23)
Taken together we have
f 0 (x) = −(f (x))2
• Since xn f (x) = 1, differentiating both sides with respect to x gives
0 = nxn−1 x−n + xn f 0 (x) = nx−1 + xn f 0 (x)
(24)
by the product rule. Solving for f 0 (x) we have
f 0 (x) = (−nx−1 )x−n = −nx−n−1
as expected.
3
(25)
2. Find the slope, that is dy/dx, of the curve
3y 2 − 2x2 = xy
(26)
at the point (1, 1).
Solution: Differentiating both sides w.r.t. x gives
6yy 0 − 4x = y + xy 0
or
y0 =
4x + y
6y − x
and so at (1, 1), substituting x = 1 and y = 1 we have the slope equal to 1.
4
(27)
(28)