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IMPLICIT DIFFERENTIATION
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Substitute 6x for y in the equation of the curve, 3x 2 x(6x) + 4(6x)2 = 141 3x 2 6x 2 + 144x 2 = 141 141x 2 = 141 x2 = 1 x = 1 Hence, ( 1, 6) is another point for which the slope of the tangent line is zero.
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16.2 If y = f (x) is a function satisfying the equation x 3 y2 2x + y3 = 36, nd a formula for the derivative dy/dx.
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Dx (x 3 y2 2x + y3 ) = Dx (36) Dx (x 3 y2 ) 2Dx (x) + Dx (y3 ) = 0 x 3 Dx (y2 ) + y2 Dx (x 3 ) 2(1) + 3y2 x 3 2y dy dx dy =0 dx dy + y2 (3x 2 ) 2 + 3y2 =0 dx (2x 3 y + 3y2 ) dy = 2 3x 2 y2 dx dy 2 3x 2 y2 = 3 dx 2x y + 3y2
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16.3 If y = f (x) is a differentiable function satisfying the equation x 2 y3 5xy2 4y = 4 and if f (3) = 2, nd the slope of the tangent line to the graph of f at the point (3, 2).
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Dx (x 2 y3 5xy2 4y) = Dx (4) x 2 (3y2 y ) + y3 (2x) 5(x(2yy ) + y2 ) 4y = 0 3x 2 y2 y + 2xy3 10xyy 5y2 4y = 0 Substitute 3 for x and 2 for y, 108y + 48 60y 20 4y = 0 44y + 28 = 0 y =
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7 Hence, the slope of the tangent line at (3, 2) is 11 .
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7 28 = 44 11
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16.4 (a) Find a formula for the slope of the tangent line to the curve x 2 xy + y2 = 12 at any point (x, y). Also, nd the coordinates of all points on the curve where the tangent line is: (b) horizontal; (c) vertical. 16.5 Consider the hyperbola 5x 2 2y2 = 130. (a) Find a formula for the slope of the tangent line to this hyperbola at (x, y). (b) For what value(s) of k will the line x 3y + k = 0 be normal to the hyperbola at a point of intersection
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IMPLICIT DIFFERENTIATION
[CHAP. 16
16.6 Find y by implicit differentiation. (a) x 2 + y2 = 25 (d) (g) x+ y=1 y2 x2 + =1 9 4 (b) x 3 = 2x + y 2x y (c) 1 1 + =1 x y x+y x y
(e) x 3 y3 = 2xy (h) y + xy3 = 2x
(f ) (7x 1)3 = 2y4 (i) x 2 =
16.7 Use implicit differentiation to nd the slope-intercept equation of the tangent line at the indicated point. x2 3 3 xy = 2 at (3, 2) 2 = 1 at 2, (a) y +y (b) 16 2 (c) (y x)2 + y3 = xy + 7 at (1, 2) (e) 4xy2 + 98 = 2x 4 y4 at (3, 2) (g) x3 y = x at (1, 1) 1 y3 (d) x 3 y3 = 7xy at (4, 2) (f ) 4x 3 xy 2y3 = 1 at (1, 1) (h) 2y = xy3 + 2x 3 3 at (1, 1)
16.8 Use implicit differentiation to nd the slope-intercept equation of the normal line at the indicated point. (a) y3 x + 2y = x 2 at (2, 1) (c) y x x y = 12 at (9, 16) (b) 2x 3 y + 2y4 x 4 = 2 at (2, 1) (d) x 2 + y2 = 25 at (3, 4) 1 7 1 x at x = 16 . [Hint: Eliminate
16.9 Use implicit differentiation to nd the slope of the tangent line to the graph of y = the radicals by squaring twice.]