2.8 Implicit Differentiation Let y fx . Then U dy f x . How can we compute dx d ¡hy ¢ ? dx Why do we need it? Example Find the equation of the tangent line to the circle x 2 y 2 4 at the point where x 1. Example Find the equation of the tangent line to the curve x 3 y 3 2xy at the point 1, 1 . Observe that d dx hy d dx hfx U U U h fx f x h y dy dx Example Suppose that y is a function of x, that is, implicitly y fx for some function f. Compute the following derivatives. d ¡y¢, d ¡y 2 ¢, d ¡ y ¢, d ¡e xy ¢, d ¡lny 2 x 2 ¢, d ¡yx¢, d ¡x 2 y 3 ¢, d ¡cos 2 xy ¢ dx dx dx dx dx dx dx dx d ¡y¢ dy dx dx d ¡y 2 ¢ 2y dy dx dx d ¡ y ¢ 1 dy 2 y dx dx d ¡e xy ¢ e xy 1 dy dx dx dy dy 1 d ¡lny 2 x 2 ¢ 2y 2x 2 2 2 y x 2 2 dx dx dx y x y x d ¡yx¢ dy x y1 y x dy dx dx dx dy dy d ¡x 2 y 3 ¢ 2xy 3 x 2 3 y 2 2xy 3 3x 2 y 2 dx dx dx dy dy d ¡cos 2 xy ¢ 2 cosxy " sinxy 1 y x " sin2xy y x dx dx dx dy Using Implicit Differentiation: dx dy Given an equation in x and y: Fx, y Gx, y , the steps for finding are: dx a. Differentiating both sides of the equation with respect to x with considering y fx implicitly. d Fx, y d Gx, y dx dx dy in terms of x and y. b. Solve dx Steps for Computing dy where x 3 y 3 2xy. Find the equation of the tangent line to the curve x 3 y 3 2xy dx at the point 1, 1 . Example Find 1 a. d ¡x 3 y 3 ¢ d ¡2xy¢ ® 3x 2 3y 2 dy 2¡1 y x dy ¢ 2y 2x dy dx dx dx dx dx b. Solve for dy . dx 3y 2 2y " 3x 2 dy dy dy dy " 2x 2y " 3x 2 ® 3y 2 " 2x 2y " 3x 2 ® dx dx dx dx 3y 2 " 2x c. Let x 1 and y 1. m 21 " 31 2 "1 "1 2 1 31 " 21 y " 1 "1 x " 1 , y "x 2 2 y1 -2 -1 1x 2 -1 -2 Example Find dy where y 5 3x 2 y 2 5x 4 12. dx a. d ¡y 5 3x 2 y 2 5x 4 ¢ d ¡12¢ ® 5y 4 dy 3¡2xy 2 x 2 2y dy ¢ 20x 3 0 dx dx dx dx b. Solve for dy . dx 5y 4 dy dy dy 6x 2 y "6xy 2 " 20x 3 ® y5y 4 6x 2 "2x3y 2 10x 2 dx dx dx "2x3y 2 10x 2 dy dx y5y 4 6x 2 Example Find the equation of the line tangent to the curve 2 x2 y 1 9 36 at the point "1, 4 2 . The equation of the tangent line at the given point is: dy at "1, 4 2 . y " 4 2 1 mx 1 where m dx 2 a. 2 2 d ¡ x 2 y ¢ d ¡1¢ ® 2x 2y dy 0 ® 2x y dy 0 36 dx 18 dx 36 9 9 dx dx 9 dy . b. Solve for dx y dy dy " 2x ® " 2x 18 " 4x y 18 dx 9 9 y dx c. 4"1 1 m " 4x y at "1, 4 2 ® m " 4 2 2 The equation of the tangent line is: y"4 2 1 x 1 . 2 Example Find the locations of all horizontal and vertical tangents of the curve: x 2 y 3 " 3y 4. a. d ¡x 2 y 3 " 3y¢ d ¡4¢ dx dx dy dy 2x 3y 2 "3 0 dx dx b. dy dy 2x "2x, 3y 2 " 3 dx dx 31 " y 2 c. Horizontal tangents: m 0 2x 0 31 " y 2 « 2x 0, x 0. When x 0, y 3 " 3y 4, solve it by a calculator: y 2. 195 823 345 The location of the horizontal tangent line is 0, 2. 20 TI-89: F2/1 - solve( solve(y^3-3y-40,y) 4. Vertical tangents: m . : 2x . 31 " y 2 « 31 " y 2 0, y 2 1, y1 y "1 When y 1, x 2 1 " 3 4, x 2 6, x o 6 When y "1, x 2 " 1 3 4, x 2 2, x o 2 Locations of vertical tangent lines are 6 ,1 , " 6 ,1 , 2 , "1 , " 2 , "1 . Example If x 2 6xy y 2 8, find y UU . a. d ¡x 2 6xy y 2 ¢ d ¡8¢ ® 2x 6¡y x dy ¢ 2y dy 0 dx dx dx dx b. Solve for 3 dy . dx dy dy dy dy x 3y 2y "2x " 6y ® 23x y "2x 3y ® " 3x y dx dx dx dx dy dy d2y . c. Compute d ¡ ¢ using the Quotient Rule and dx dx dx dx 2 dy dy 1 3 3x y " x 3y 3 x 3y d2y dx dx d ¡" ¢ " dx 3x y dx 2 3x y 2 dy dy 3x y 33x y " 3x 3y " x 3y dx dx " 3x y 2 dy 9x 3y " x " 3y " 3x " 9y 3x y dx " 3x y 2 x 3y dy 8x " " y 8x " 8y 3x y dx " " 3x y 2 3x y 2 8¡x3x y " x 3y " y3x y ¢ 3x y 3 8¡3x 2 xy " x " 3y " 3xy y 2 ¢ 8¡3x 2 " x " 3y " 2xy y 2 ¢ 3x y 3 3x y 3 6x 4
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