MA1S11 Tutorial Sheet 7, Solutions1
24 - 27 November 2015
Questions
The numbers in brackets give the numbers of marks available for the question.
1. (4) Determine the derivative of the inverse function y = f −1 (x), by implicit differentiation of the equation
f (y) = x
(1)
√
Then use the result to obtain the derivatives of f −1 (x) for f (x) = x (x ≥ 0) and
f (x) = cos x (for x ∈ [0, π]).
(Hint: for the inverse cosine function you’ll need the identity cos2 x + sin2 x = 1)
Solution: We differentiate both sides of the equation w.r.t. x to obtain
f ′ (y)
dy
=1
dx
⇔
dy
1
= ′ ,
dx
f (y)
(2)
provided f ′ (y) =
6 0, which we assume here. In the √
case f (x) =
√
equation reads f (y) = y = x. One has f ′ (x) = 1/(2 x) and
√
x, the original
1
dy
√
=
√ = 2 y = 2x
dx
1/(2 y)
(3)
For f (x) = cos x we set have cos(y) = x and cos′ (x) = − sin(x), so
dy
1
1
−1
−1
,
= ′
=
=p
=√
dx
f (y)
− sin y
1 − x2
1 − cos2 y
so that we conclude
−1
d
.
cos−1 (x) = √
dx
1 − x2
(4)
(5)
2. (4) Determine the zeros, the relative maxima and minima, the regions of concavity
up/down and the end behaviour for x → ±∞ for
f (x) = (x + 1)2 (x − 1) = x3 + x2 − x − 1
(6)
Then draw the graph of f using all the gathered information. Solution:
Using the factorised form we read of the zeros of f as x = 1, −1. The function is
polynomial in x and therefore differentiable everywhere. Critical points are obtained
by solving f ′ (x) = 0,
f ′ (x) = 3x2 + 2x − 1 = (3x − 1)(x + 1)
1
⇒
x = 1/3, −1.
Stefan Sint, sint@maths.tcd.ie, see also http://www.maths.tcd.ie/~sint/MA1S11.html
1
(7)
To find out about the regions of concavity we differentiate again and obtain
f ′′ (x) = 6x + 2.
(8)
Thus f ′′ (x) is positive for x > −1/3 and negative for x < −1/3; x = −1/3 is the
point of inflection dividing the regions concave down (x < −1/3) and concave up
(x > −1/3). To determine which of the stationary points are maxima, minima or
neither we use the second derivative test to find f ′′ (1/3) = 4 > 0 and f ′′ (−1) =
−4 < 0, so that a relative maximum is obtained at x = −1 and a relative minimum
at x = 1/3. Finally the end behaviour is obtained as
lim f (x) = +∞,
lim f (x) = −∞
x→∞
(9)
x→−∞
as the leading power is x3 . Hence we obtain the following graph:
3
2
1
K2
K1
x
0
1
K1
K2
K3
Figure 1: graph of f (x) in question 2.
Extra Questions
The questions are extra; you don’t need to do them in the tutorial class.
1. (4) Find dy/dx and d2 y/dx2 for
sin xy = 0
2
(10)
by implict differentiation. Compare the result to the direct differentiation of the
1-parameter family of functions y = fn (x) (n = 0, ±1, ±2, . . .) obtained as solutions
from eq. (10).
Solution:
Differentiating across we get
y cos(xy) + x
dy
cos(xy) = 0.
dx
(11)
Note that cos(xy) 6= 0 as xy must be solution of sin(xy) = 0, so we can divide by
cos(xy) and obtain
dy
=0
(12)
y+x
dx
and
y
dy
=−
(13)
dx
x
Now, differentiating equation (12) again gives
dy dy
d2 y
+
+x 2 =0
dx dx
dx
(14)
so
2 dy
2y
d2 y
=−
= 2
2
dx
x dx
x
Now, each of the zeros of sin(xy) can be used to define a function
y = fn (x) =
nπ
,
x
n = 0, ±1, ±2, . . .
(15)
(16)
Differentiating we obtain
fn′ (x) =
−nπ
y
=
−
,
x2
x
fn′′ (x) =
2nπ
2y
=
x3
x2
(17)
so, indeed each of the function satisfies the equations derived above by implicit differentiation. In the case n = 0 we simply have f0 = f0′ = f0′′ = 0.
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