12.25
Find a formula for the nth derivative of y = 1 lx(\ - x).
I Observe that y = l/x+ 1/(1-*). Now, the nth derivative of l/x is easily seen to be (-!)"(
and that of !/(*-!) to be («!)(!- x)( "
+ l \ Hence, /"' = (n!)[(-l)"/*"
+ 1 + 1/(1 - Jt)"
+ I )].
I /'(*)
= 2* if x>0 and f'(x)=-2x if x<0. Direct computation by the A-definition, shows that
/'(0) = 0. Hence, f'(x) = 2\x\ for all x. Since \x\ is not differentiable at *=0, /"(O) cannot exist.
12.27
Consider the circles C,: (x - a)
2 + y
2 = 8 and C 2 : (x + a)
2 + y
2 =8. Determine the value of |a| so that C,
and C, intersect at right angles.
I Solving the equations for C, and C 2 simultaneously, we find (x — a)
2 = (x + a)
2 , and, therefore, x = 0.
Hence, y = ±V8- a
2 . OnC 1; 2(x l - a) + 2y l y' l = 0, so at the intersection points, y l y' l = a. (Here, the
subscript indicates values on C,.) On C 2 , 2(x 2 + a) + 1y^y\ = 0, so at the intersection points, y 2 y' 2 = —a.
Hence, multiplying these equations at the intersection points, y\y\y-iy'i
= ~<*
L - At these points, y\—y->
= y\
hence, y
2 y\y' 2
= ~o
2 . Since C, and C 2 are supposed to be perpendicular at the intersection points, their tangent
lines are perpendicular, and, therefore, the product y[y' 2 of the slopes of their tangent lines must be -1. Hence,
-y
2 =-a
2 , y
2 = a
2 . But y
2 = 8 — a
2 at the intersection point. So, a
2 = 8 — a
2 , a
2 =4, |a|=2.
12.28
Show that the curves C, : 9y - 6x + y* + x
} y = 0 and C,: Wy + I5x + x
2 - xy
3 = 0 intersect at right angles
at the origin.
I On C,, 9y' -6 + 4y
3 y' + x
3 y' + 3x
2 y = 0. At the origin (0,0), 9/-6 = 0, or /=§. On C,,
10/ + 15 + 2x-y
3 -3xy
2 y' = 0. At the origin, 10>>' + 15 = 0, or / = -§. Since the values of y' on C,
and C 2 at the origin are negative reciprocals of each other, the tangent lines of C\ and C 2 are perpendicular at the
origin.
In Problems 12.29 to 12.35, calculate the second derivative y".
12.29
12.30
12.32 y = (x + l)(x - 3)
3
.
I Here it is simplest to use the product rule (uv)" = u"v + 2u'v' + uv". Then y" = (0)(x - 3)
} + 2(l)[3(x3)
2
] + (x + 1)[6(* - 3)] = 12(jr - 3)(* - 1).
12.26
Consider the function f(x) defined by
if
if
Show that /"(O) does not exist.
78
CHAPTER 12
12.31