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Unformatted text preview: [Hand back homework and hand out practice test solutions.] What’s the main idea of section 2.6? ... Implicit differentiation: To differentiate y with respect to x , you don’t always need to write y explicitly in the form f ( x ); it can be enough to write an algebraic relationship between x and y . The tangent to the unit circle at the point P =( x , y ) has slope – x / y . One way to prove this is with elementary geometry (use the fact that the tangent line is perpendicular to the radius OP , and the fact that the radius OP has slope y / x , and the fact that two lines are perpendicular iff the product of their slopes is –1). Another way is to write the top half of the circle as the graph of the function sqrt(1– x 2 ) and take derivatives, and then do the same for the bottom half of the circle using the function –sqrt(1– x 2 ). Here’s how we do it with implicit differentiation: x 2 + y 2 = 1 (an algebraic relationship between x and y ). Differentiate with respect to x and solve for y ′ in terms of x and y : 2 x + 2 yy ′ = 0 2 yy ′ = –2 x y ′ = (–2 x )/(2 y ) = – x / y Note that this works both for the “positive branch” of the multivalued function y = ± sqrt(1– x 2 ) (i.e. y = sqrt(1– x 2 )) as well as the “negative branch” ( y = –sqrt(1– x 2 ))....
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 Fall '11
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 Calculus, Derivative, Implicit Differentiation, Continuous function

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