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# lec12 - 12 Implicit Dierentiation P K Lamm Lecture...

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12. Implicit Differentiation P. K. Lamm 10/06/11 (18:56) p. 1 / 6 Lecture Notes: 12. Implicit Differentiation These classnotes are intended to be supplementary to the textbook and are necessarily limited by the time allotted for classes. For full and precise statements of definitions and theorems, as well as material covering other topics and examples, please consult the textbook. 1. Implicit Equations and Implicit Differentiation Definition: An implicit equation in x and y is one in which neither x nor y is given explicitly. Example 1.1: Find the tangent line to the curve x 2 + y 2 = 1 at the point 3 5 , 4 5 . The equation x 2 + y 2 = 1 is an implicit equation in x and y and its solution is the set of all ( x, y ) which satisfy the equation, i.e., { ( x, y ) | x 2 + y 2 = 1 } . It is not clear whether y should be the dependent variable and x the independent variable or vice versa. For now let’s take y to be the dependent variable and solve for y in terms of x : y 2 = 1 - x 2 y = ± 1 - x 2 . We thus get two functional representations for y : y = + 1 - x 2 gives an expression for y on the upper half of the unit circle, y = - 1 - x 2 gives an expression for y on the lower half of the unit circle. To compute y 0 ( x ) at 3 5 , 4 5 , we would need the expression for y on the upper half of the unit circle (since 4 / 5 > 0.) Thus we compute y 0 ( x ) = d dx 1 - x 2 = d dx (1 - x 2 ) 1 / 2 = 1 2 (1 - x 2 ) - 1 / 2 ( - 2 x ) = - x 1 - x 2 where we have used the Chain Rule in addition to a rule for differentiating fractional powers of x that we will actually prove in Section 2 below. Thus, when x = 3 / 5, y 0 (3 / 5) = - 3 / 5 q 1 - (3 / 5) 2 = - 3 / 5 q 1 - (9 / 25) = - (3 / 5) 4 / 5 = - 3 4 . 1

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12. Implicit Differentiation P. K. Lamm 10/06/11 (18:56) p. 2 / 6 Thus the tangent line to the curve at 3 5 , 4 5 is given by y - 4 5 = - 3 4 x - 3 5 .
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lec12 - 12 Implicit Dierentiation P K Lamm Lecture...

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