12.20
Find y" on the parabola y
2 = 4px.
I By implicit differentiation, 2yy'=4p, yy'= 2p. Differentiating again and using the product rule, yy" +
y'y'=0. Multiply both sides by y
2 : y*y" + y
2 (y')
2 =Q- However, since yy' = 2p, y
2 (y')
2 =
4p
2
. So, y*y" + 4p
2 = 0, and, therefore, y"=—4p
2 /y
3 .
12.21
Find a general formula for y" on the curve x" + y" = a".
By implicit differentiation, nx"~l + ny"~ly'=Q. that is, (*) x"~l + y"~'y'=0. Hence, y"~ly' =
—x"'', and, therefore, squaring, (**) y2"~2(y')2 - x2"'2. Now differentiate (*) and multiply by y":
(n - l)jt"-y + [y2"~V" + (« - 1)2"~V)2] -0. Use (**) to replace y2"~2(y')2 by x2'-2: (n - l)x"-2y" +
[y2"-ly" + (n - I)x2"~2] = 0. Hence,"~ly" = - (n - I)(x2"~2 + *"~2y") = ~(« - l)x"~2(x" + y") =
2
- (n - l)x"~
2 a". Thus, y" = -(n - I)a"x"~
2 /y
2 "~\ (Check this formula in the special case of Problem
12.11).
12.22
Find y" on the curve x
1 '
2 + y
1 '
2 = a"
2 .
I Use the formula obtained in Problem 12.21 in the special case n=k: y" = -(- ka
l>2 x~*'
2 )/y° = ±a
ll2 /x
3 '
2 .
12.23
Find the 10th and llth derivatives of the function f(x) = ^
10 - Ux
7 + 3>x' + 2x* -x + 2.
I By the 10th differentiation, the offspring of all the terms except *
10 have been reduced to 0. The successive
offspring of x
10 are lOx
9
, 9 • 10x
8
, 8 • 9 • 10x
7
,
1-2-3
10. Thus, the 10th derivative is 10!. The llth
derivative is 0.
2.24
For the curve y
3 = x
2 , calculate y' (a) by implicit differentiation and (b) by first solving for y and then
differentiating. Show that the two results agree.
I (a) 3y
2 y'=2x. Hence, y' =2x/3y
2 . (b)y = x
2 '\ So, y' = lx~"\ Observe that, since y
2 = x"\
the two answers are the same.
HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION
77
Use implicit differentiation. (*) 2x + 2(xy' + y) + 6yy' = 0. When y = l, the original equation yields
x
2 + 2x + 3 = 2, x
2 + 2x + I = 0, (x+l)
2 = 0, * + l=0, * = -!. Substitute -1 for x and 1 for y in (*),
which results in —2 + 2(—y' + l) + 6y'=0; so, y'=0 when y = l. To find y", first simplify (*) to
x + xy' + y + 3yy' = 0, and then differentiate implicitly to get 1 + (xy" + y') + y' + 3(yy" + y'y') = 0. In this
equation, substitute —1 for ;c, 1 for y, and 0 for y', which results in 1 - y" + 3y" = 0, y" = -1.
12.16
Find the slope of the tangent line to the graph of y = x + cos xy at (0,1).
Differentiate implicitly to get y' = 1 - [sin xy • (xy' + y)]. Replace x by 0 and y by 1. y' = 1 -
[sin (0) • 1] = 1 - 0 = 1. Thus, the tangent line has slope 1.
12.17 If cosy = x, find/.
Differentiate implicitly: (-sin y)y' = 1. Hence, y' = —l/(siny) =
12.18
Find an equation of the tangent line to the curve 1 + 16* y = tan (x - 2y) at the point (Tr/4, 0).
Differentiating implicitly, I6(x
2 y' + 2xy) = [sec
2
(.v - 2y)](l -2y'). Substituting w/4 for x and 0 for y,
16(ir
2 /16)(y') = [sec
2 (ir/4)](l-2/). Since cos (77/4) = V3/2, sec
2 (77/4) = 2. Thus, Try' = 2(1 - 2y').
Hence, y' =21(17* + 4), which is the slope of the tangent line. A point-slope equation of the tangent line is
y = [2/(7r
2 + 4)](x-7r/4).
12.19
Evaluate y" on the ellipse b
2 x
2 + ary
2 = a
2 b
2 .
Use implicit differentiation to get 2b
2 x + 2a
2 yy' = 0, y' = -(b
2 /a
2 )(x/y).
Now differentiate by the
quotient rule.