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Implicit Differentiation Lesson 3.6 1 2 Introduction Consider an equation involving both x and y: 2 2 x y 49 This equation implicitly defines a function in x It could be defined explicitly y x 49 2 (where x 7) Differentiate Differentiate both sides of the equation each term one at a time use the chain rule for terms containing y For x y 49 we get 2 2 dy 2x 2 y 0 dx Now solve for dy/dx 3 Differentiate Then dy 2x 2 y 0 gives us dx dy 2 x x dx 2 y y We can replace the y in the results with the explicit value of y as needed This gives us dy x the slope on the 2 dx x 49 curve for any legal value of x 4 Slope of a Tangent Line Given x3 + y3 = y + 21 find the slope of the tangent at (3,-2) 3x2 +3y2y’ = y’ 2 3 x Solve for y’ y' 2 1 3y Substitute x = 3, y = -2 27 slope 11 5 6 Second Derivative Given x2 –y2 = 49 x y' y y’ =?? y’’ = Substitute d y y x y' 2 2 dx y 2 7 Derivatives of Inverse Trig Functions See Theorem 3.11, pg 182 Try this y = cos-1(4x – 3) dy 1 4 dx 1 (4 x 3) Exponential & Log Functions Given y = bx where b > 0, a constant dy x ln b b dx Given y = logbx 1 y' ln b x Note: this is a constant 8 9 Using Logarithmic Differentiation Given y ( x 1) ( x 3) 18 10 3 7 8 Take the log of both sides, simplify Now differentiate both sides with respect to x, solve for dy/dx 10 Implicit Differentiation on the TI Calculator We can declare a function which will do implicit differentiation: Usage: Assignment Lesson 3.6 Page 186 5, 7, 9, 13, 15, 17, 19, 21, 29, 30, 33, 35, 39, 41, 51, 57 11