1S1 Tutorial Sheet 6: Solutions1
7 February 2007
Questions
1. (4) Find the first four terms in the Taylor expansion of ln(x) about x = 1.
Solution: So we are interest in the expansion of ln (1 + h) in h:
1 2 d2 ln x d ln x + h
ln (1 + h) = ln (1) + h
dx
2
dx2 x=1
x=1
3
4
1 d ln x 1 4 d ln x + h3
h
+
+...
6
dx3 x=1 24
dx4 x=1
(1)
Now
d ln x
dx
d2 ln x
dx2
d3 ln x
dx3
4
d ln x
dx4
so
=
1
x
= −
=
1
x2
2
x3
= −
6
x3
1
1
1
ln (1 + h) = h − h2 + h3 − h4 + . . .
2
3
4
(2)
(3)
2. (2) Use implicite differentiation to find dy/dx where
ln x + 2xy + y 2 = 0
(4)
Solution: Differenciating across we get
1
dy
dy
+ 2y + 2x + 2y
x
dx
dx
(5)
dy
1/x + 2y
1 + 2xy
=−
=−
dx
1 + 2y
x + 2xy
(6)
so, solving for dy/dx we get
1
Conor Houghton, houghton@maths.tcd.ie, see also http://www.maths.tcd.ie/~houghton/1S1
1
3. (2) Find
x2 − 1
x→1 x2 − x
both by factorizing and by l’Hôpital’s rule.
lim
(7)
Solution: First by factorizing:
(x − 1)(x + 1)
x+1
x2 − 1
= lim
= lim
=2
2
x→1
x→1
x→1 x − x
x(x − 1)
x
lim
(8)
and, we can see from the above that both numerator and denominator vanish at x = 1 so
we can use l’Hopital’s rule:
x2 − 1
2x
= lim
=1
2
x→1 x − x
x→1 2x − 1
lim
(9)
Extra Questions
1. Find df /dx where f (x) = cos−1 x.
Solution: So, say y = cos−1 x and hence x = cos y. Now differentiate
dy
dx
(10)
1
dy
=−
dx
sin y
(11)
1 = − sin y
and hence
p
and now it would be good to rewrite this in terms of x; sin y = ± 1 − cos2 y and
choosing the positive square root, making the sine positive, as is appropriate for the
interval zero to π.
1
dy
= −√
(12)
dx
1 − x2
where we have used cos y = cos (cos−1 x) = x.
2. Find dy/dx where ln y + ln x = exp xy.
Solution: Differentiating
1 dy 1
dy
+ = yexy + x exp xy
y dx x
dx
(13)
yexy − x1
dy
xy 2 exy − y
= 1
=
dx
x − xyexy
− xexy
y
(14)
so, solving for dy/dx
2
3. Differentiate ln cos x.
Solution: This is done using the chain chain rule
d
sin x
ln cos x = −
= − tan x
dx
cos x
(15)
4. Differentiate ln x3 both by the chain rule and by ln x3 = 3 ln x.
Solution:
So, by the chain rule we get
and
5. Find
3x2
3
d
ln x3 = 3 =
dx
x
x
(16)
d
3
3 ln x =
dx
x
(17)
ex − 1
x→0 sin x
(18)
lim
Solution: So the top and bottom are both zero at the limit point x = 0, so, differentiating top and bottom
ex
ex − 1
= lim
=1
(19)
lim
x→0 cos x
x→0 sin x
6. Find
ln x
x→1 1 − x
lim
(20)
Solution: Again the top and bottom are both zero at x = 1 so
1
ln x
= − lim = −1
x→1 x
x→1 1 − x
(21)
ln x
x→1 exp (x − 1)
(22)
lim
7. Find
lim
Solution: Trick question, the denominator isn’t zero at x = 1, in fact you kiust
substitute
ln x
=0
(23)
lim
x→1 exp (x − 1)
3
8. Find
x2 − 1
x→1 x2 − 2x + 1
lim
(24)
Solution: Again, and bottom are zero at x = 1, so differentiate
x2 − 1
2x
lim
= lim
x→1 x2 − 2x + 1
x→1 2x − 2
(25)
so now the bottom is zero but the top isn’t and the sign is different for x approaching
from above and x approaching from below, so the limit is not defined.
4