This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: [Hand back homework and hand out practice test solutions.] Whats the main idea of section 2.6? ... Implicit differentiation: To differentiate y with respect to x , you dont 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 ). Heres 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 ))....
View Full Document