HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION
12.42
Find the equations of the tangent lines to the ellipse 9x
2 + 16y
2 = 52 that are parallel to the line 9x-8y = l.
By implicit differentiation, I8x + J>2yy' = 0, y' = -(9;t/16>'). The slope of 9x — 8.y = 1 is \. Hence,
for the tangent line to be parallel to 9x-8y=\, we must have -(9x/l6y) = |, -x = 2y. Substituting in
the equation of the ellipse, we obtain 9(4y
2 ) + I6y
2 = 52, 52y~ = 52, y
2 = 1, y = ±l. Since x = -2y,
the points of tangency are (—2,1) and (2, —1). Hence, the required equations are y - 1 = g(x + 2) and
y + l=ti(x-2), or 9*-By =-26 and 9x-Sy = 26.
79
12.33
12.34 x
2 -y
2 = l.
12.35
12.36
Find all derivatives of y = 2x
2 + x — l + l/x.
12.37
At the point (1,2) of the curve x
2 - xy + y
2 = 3, find the rate of change with respect to * of the slope of the
tangent line to the curve.
and,
for «>4, y
( -
) = (-l)"(n!)j C -
( "
+ I> .
By implicit differentiation, (*) 2x - (xy
1 + y) + 2yy' = 0. Substitution of (1,2) for (x, y) yields^ 2(y
1 + 2) + 4y' = 0, / = 0. Implicit differentiation of (*) yields 2 - (xy" + y' + y') + 2yy" + 2(/)
2 = 0,
Substitution of (1,2) for (x, y), taking into account that y' = 0 at (1, 2), yields 2 +(4-!)>>" = 0, /'=-§.
This is the rate of change of the slope y' of the tangent line.
In Problems 12.38 to 12.41, use implicit differentiation to find y'.
12.38
tan xy = y.
(sec
2 xy) • (xy' + y) = y'.
Note that
sec
2 xy — 1 + tan
2 xy = 1 + y'.
Hence,
(1 + y
2 )(xy' + y) = y',
y'[x(i + r)-1] = -Xi + y2), y' = y(i + y2)/[i -*(i + y2)}12.39
sec
2 y + cot
2 x = 3.
(2 sec y)(sec y tan y)y' + (2 cot Jt)(-csc
2 x) = 0, y' = cot x esc
2 .v/sec
2 y tan y.
12.40
tan
2 (y+ l) = 3sin.*:.
tan
2 (y + l) + l = 3sinx + l, the answer can also be written as y' =
2tan(y + l)sec
2 (y + l)y' = 3 cos*. Hence,
Since sec
2 (y + 1) =
12.41
y = tan
2 C*r + y).
Note that sec
2 (x + y) = tan
2 (x + y) + 1 = y + 1. y' = 2 tan (x + y) sec
2 (x + y)(l + y') = 2 tan (x + y) x
(y + l)(l+y'). So, y'[l-2tan(x + y)(y + l)] = 2tan(;c + y)(y + 1),
2*-2yy'=0, x-yy'=0,
y =2x
2 + x-l + x~\ y'=4x + l-x'
2 , y" = 4 + 2 X -\ y'" = -(3 • 2)x~\ >'
(4> = (4-3 • 2)x'\
12.42
Find the equations of the tangent lines to the ellipse 9x
2 + 16y
2 = 52 that are parallel to the line 9x-8y = l.
By implicit differentiation, I8x + J>2yy' = 0, y' = -(9;t/16>'). The slope of 9x — 8.y = 1 is \. Hence,
for the tangent line to be parallel to 9x-8y=\, we must have -(9x/l6y) = |, -x = 2y. Substituting in
the equation of the ellipse, we obtain 9(4y
2 ) + I6y
2 = 52, 52y~ = 52, y
2 = 1, y = ±l. Since x = -2y,
the points of tangency are (—2,1) and (2, —1). Hence, the required equations are y - 1 = g(x + 2) and
y + l=ti(x-2), or 9*-By =-26 and 9x-Sy = 26.
79
12.33
12.34 x
2 -y
2 = l.
12.35
12.36
Find all derivatives of y = 2x
2 + x — l + l/x.
12.37
At the point (1,2) of the curve x
2 - xy + y
2 = 3, find the rate of change with respect to * of the slope of the
tangent line to the curve.
and,
for «>4, y
( -
) = (-l)"(n!)j C -
( "
+ I> .
By implicit differentiation, (*) 2x - (xy
1 + y) + 2yy' = 0. Substitution of (1,2) for (x, y) yields^ 2(y
1 + 2) + 4y' = 0, / = 0. Implicit differentiation of (*) yields 2 - (xy" + y' + y') + 2yy" + 2(/)
2 = 0,
Substitution of (1,2) for (x, y), taking into account that y' = 0 at (1, 2), yields 2 +(4-!)>>" = 0, /'=-§.
This is the rate of change of the slope y' of the tangent line.
In Problems 12.38 to 12.41, use implicit differentiation to find y'.
12.38
tan xy = y.
(sec
2 xy) • (xy' + y) = y'.
Note that
sec
2 xy — 1 + tan
2 xy = 1 + y'.
Hence,
(1 + y
2 )(xy' + y) = y',
y'[x(i + r)-1] = -Xi + y2), y' = y(i + y2)/[i -*(i + y2)}12.39
sec
2 y + cot
2 x = 3.
(2 sec y)(sec y tan y)y' + (2 cot Jt)(-csc
2 x) = 0, y' = cot x esc
2 .v/sec
2 y tan y.
12.40
tan
2 (y+ l) = 3sin.*:.
tan
2 (y + l) + l = 3sinx + l, the answer can also be written as y' =
2tan(y + l)sec
2 (y + l)y' = 3 cos*. Hence,
Since sec
2 (y + 1) =
12.41
y = tan
2 C*r + y).
Note that sec
2 (x + y) = tan
2 (x + y) + 1 = y + 1. y' = 2 tan (x + y) sec
2 (x + y)(l + y') = 2 tan (x + y) x
(y + l)(l+y'). So, y'[l-2tan(x + y)(y + l)] = 2tan(;c + y)(y + 1),
2*-2yy'=0, x-yy'=0,
y =2x
2 + x-l + x~\ y'=4x + l-x'
2 , y" = 4 + 2 X -\ y'" = -(3 • 2)x~\ >'
(4> = (4-3 • 2)x'\