MA1S11 Tutorial Sheet 6, Solutions1
17 November 2015
Questions
The numbers in brackets give the numbers of marks available for the question.
1. (2) 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
(1)
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
(2)
d n
x = −x−2n nxn−1 = −nx−n−1
dx
(3)
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)
(4)
by the product rule. Solving for f 0 (x) we have
f 0 (x) = (−nx−1 )x−n = −nx−n−1
as expected.
1
Stefan Sint, sint@maths.tcd.ie, see also http://www.maths.tcd.ie/~sint/MA1S11.html
1
(5)
2. (4) Differentiate the following functions with respect to x:
√
sin x
x3 cos 3x,
sin (cos x),
,
−x2 cos x
2
cos x
(6)
Solution: So, differentiating x3 gives 3x2 , and differentiating cos 3x gives −3 sin 3x
using the chain rule. Then, by the product rule,
d 3
x cos (3x) = 3x2 {cos(3x) − x sin(3x)}.
(7)
dx
Again use the chain rule with f (x) = sin x and g(x) = cos x,
d
sin (cos x) = − cos(cos x) sin x.
(8)
dx
In the next one we have an example of the quotient rule with f (x) = sin(x) and
g(x) = cos2 x, and we need the chain rule to differentiate g,
g 0 (x) =
d
cos2 x = −2 cos x sin x.
dx
(9)
Taken together this gives
d
d
cos2 x dx
sin x − sin x dx
cos2 x
cos2 x + 2 sin2 x
d sin x
=
=
(10)
dx cos2 x
cos4 x
cos3 x
For √
the last one we need again product and chain rules with f (x) = −x2 and g(x) =
cos x we have
√
√
√
d
d
(−x2 cos x) = −2x cos x − x2 cos x.
(11)
dx
dx
The chain rule implies
√
√
1
d
cos x = − sin( x) √ ,
(12)
dx
2 x
so that
√
√
√
d
1
(−x2 ) cos x = −2x cos x + x3/2 sin( x).
(13)
dx
2
Solution:
3. (2) Find the slope, that is dy/dx, of the curve
3y 2 − 2x2 = xy
(14)
at the point (1, 1).
Solution: Differentiating both sides w.r.t. x gives
6yy 0 − 4x = y + xy 0
(15)
or
4x + y
6y − x
and so at (1, 1), substituting x = 1 and y = 1 we have the slope equal to 1.
y0 =
2
(16)
Extra Questions
The questions are extra; you don’t need to do them in the tutorial class.
1. Find dy/dx for
3yx6 − 4x2 + 6 sin y 4 = 0
(17)
d
3yx6 − 4x2 + 6 sin y 4 = 0
dx
(18)
d
d
d 6
yx − 4 x2 + 6 sin y 4 = 0
dx
dx
dx
(19)
d 6
dy
yx = 6x5 y + x6
dx
dx
(20)
d sin y 4
dy
= 4y 3 cos(y 4 )
dx
dx
(21)
18x5 y + 3x6 y 0 − 8x + 24y 3 y 0 cos y 4 = 0
(22)
Solution:
Differentiate across so
giving
3
Using the product rule
and the chain rule
and hence
giving
y0 =
8x − 18x5 y
3x6 + 24y 3 cos y 4
(23)
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
(24)
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
(25)
so that we get
(f /g)0 = (f h)0 = f 0 h + f h0 = f 0 /g + f (−g 0 /g 2 ) =
i.e. the quotient rule.
3
f 0g − f g0
,
g2
(26)