This involves a sophisticated use of implicit differentiation.
Without a graphing calculator the graph of sin y = xy will
be hard to draw. But let us imagine that a graph was there, and that we want to find
the slope and the concavity of that graph at a certain point, for example at (0, π ). By
implicit differentiation, regarding y as an implicit function of x :
( cos y) y ′ = y + x y ′
(1)
or
( cos y - x ) y ′ = y
(2)
Hence
y ′ = y ⁄ (cos y - x ) (3)
In particular, y′( 0 ) = π ⁄ cos π - 0 = - π is the slope of the tangent line at ( 0, π ) .
The tangent line itself therefore has equation y - π = - π ( x - 0 ) , or y = π - π x .
Differentiating (3) once more with respect to x gives
(cos y − x )y ' - y(- (siny)y ' - 1)
y″=
(4)
(cos y − x) 2
using (2), (3) to simplify (4) one finally gets
2 y (cos y − x) + y 2 sin y
y″ =
(cos y − x) 3
At x = 0, y = π , one has
2π (cos π − 0) + π 2 sin π
y″ (0) =
= 2π
(cos π − 0) 3
Since 2 π > 0 the graph of sin y = xy at ( 0, π ) is concave up. In the vicinity of ( 0, π )
the graph should be drawn above the line y = π − πx .
§4.6, p.268, #21
Given z = g(y), y = f (x). So z = g ( f (x) ), and by the chain
§ 4.6, p.268, #85
rule
d 2z
dx 2
Now
dz dy
dz
=
. Applying Dx to both sides, we get
dx
dy dx
=
d ⎛ dz ⎞
⎜ ⎟
dx ⎜⎝ dy ⎟⎠
d ⎛ dz ⎞
⎜ ⎟
dx ⎜⎝ dy ⎟⎠
·
dy dz d 2 y
+
·
dx dy dx 2
(1)
d ⎛ dz ⎞ dy
d 2 z dy
⎜⎜ ⎟⎟ ·
=
·
. Substituting this into
dy ⎝ dy ⎠ dx
dy 2 dx
the box in (1) will give the final answer.
=
§ 10.4, p.715, #28
By standard formula
x2
x4
x6
cos x = 1 +
+ ·····
2 ! 4 ! 6!
substituting x by 2x gives
4x 2
16 x 4
64 x 6
+
+ ·····
cos 2x = 1 2!
4!
6!
1
1
If we multiply ( 1 ) by and then add
to the result,
2
2
1
1
2x 2
8x 4
32 x 6
we get
cos 2x = 0 +
+ ·····
2
2
2!
4!
6!
(1)
(2)
The left hand side equals f (x) = sin 2 x , so (2) is the Maclaurin series for sin 2 x .
§ 10.4, p.715, #29
Let f (x) = ℓn ( 1+x ) . Repeated differentiation gives
2 !
−1
−3 !
1
f ′ (x) =
, f ′″ (x) =
, f (iv) (x) =
٠
, f ″ (x) =
2
3
1+ x
(1 + x )
(1 + x)
(1 + x) 4
(−1) n −1 (n − 1) !
(1 + x) n
It follows that f (0) = 0, f ′ (0) = 1 , f ″ (0) = -1 , f ″′ (0) = 2 ! ,
٠٠٠٠٠ and in general f (n) (x) =
٠٠٠٠٠ f (n) (00 = (-1)n-1 (n-1) !
The Taylor formula for x near a = 0 is
f ′′ (0) 2
f ( n ) (0) n
f (x) = f (0) + f ′ (0) x +
x + ٠٠٠٠ +
x + ٠٠٠٠
2!
n !
x2 x3
f ( n ) ( 0) ! n
So ℓn ( 1 + x ) = 0 + x +
± ٠٠٠٠ +
x ± ٠٠٠٠
2
3
n !
(−1) n −1
Note that the coefficient of xn simplifies to
[ Answer ]
n
§ 10.4, Extra question 1
To get the Maclaurin series for x3 ℓn ( 1 + x ),
just multiply x3 to the answer of p.175, # 29.