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Unformatted text preview: Math 1a Implicit Differentiation Fall, 2009 ’ & $ % Often functions are defined explicitly , like y = x 3 e x or y = cos( x +1). But sometimes functions are defined implicitly , like the circle x 2 + y 2 = 1. This is really two functions y = √ 1 x 2 and y = √ 1 x 2 . Of course, sometimes the explicit functions are not so easy to find, such as x 3 + y 3 = 6 xy or x + e xy = y 2 . We can still differentiate by assuming that y is a function of x . 1 Find dy dx in terms of x and y for each of the following implicitlydefined functions. (a) x 3 + 3 x + y = e y (b) sin( x + y ) = y x (c) x 2 y = cos 2 ( xy ) (d) tan( x/y ) = y 2 x 2 2 This is convenient as we can then implicitly define inverse functions and compute their deriva tives. In this problem we’ll find the derivative of sin 1 ( x ) (also written arcsin( x )), the inverse of the sine function. (a) Rewrite y = sin 1 ( x ) as sin( y ) = x , then differentiate this function. (b) Solve for y in terms of only x , not in terms of y . (You may have to write cos( y ) in terms of sin( y ). Remember that sin 2 ( θ ) + cos 2 ( θ ) = 1 for every θ .) (c) What is d dx ( sin 1 ( x ) ) ? 3 Repeat the previous problem to find the derivatives of the following inverse functions: (a) cos 1 ( x ) = arccos( x ) (b) tan 1 ( x ) = arctan( x ) 4 Use your answers to the previous problems to compute the following derivatives: (a) 1 cos 1 ( x ) (b) √ x arctan( x ) 5 (a) Find the derivative dy dx of the functions xy = c , where c 6 = 0 is an unknown constant. (This is a family of hyperbolas.) (b) Find the derivative dy dx of the functions x 2 y 2 = k , where k 6 = 0 is an unknown constant. (This is also a family of hyperbolas, but now the asymptotes are y = ± x .) (c) At the point ( x,y ), notice that the curve through this point in each of these families are perpendicular. These families are called orthogonal trajectories . 6 Find k so that y = x k is an orthogonal trajectory for the ellipse 4 x 2 + y 2 = 1. Hint: x k 1 = y/x . Implicit Differentiation – Answers / Solutions 1 (a) If we differentiate x 3 + 3 x + y = e y with respect to x , we get 3 x 2 + 3 + y = e y y or dy dx = 3( x 2 + 1) e y 1 . (b) Differentiating sin( x + y ) = y x with respect to x gives us cos( x + y ) 1 + dy dx = dy dx 1 or dy dx = 1 + cos( x + y ) 1 cos( x + y ) . (c) Differentiating x 2 y = cos 2 ( xy ) with respect to x gives us 2 xy + x 2 dy dx = 2 cos( xy ) · d dx cos( xy ) = 2 cos( xy ) ( sin( xy )) · d dx ( xy ) = 2 sin( xy ) cos( xy ) y + x dy dx . Solving for dy dx , we get dy dx = 2 xy + 2 y sin( xy ) cos( xy ) x 2 + 2 x sin( xy ) cos( xy ) . (d) Differentiating tan( x/y ) = y 2 x 2 with respect to x , we get sec 2 ( x/y ) · d dx x y = 2 y dy dx 2 x or sec 2 ( x/y ) · y x dy dx y 2 = 2 y dy dx 2 x....
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This note was uploaded on 03/27/2012 for the course MATH 1a taught by Professor Benedictgross during the Fall '09 term at Harvard.
 Fall '09
 BenedictGross
 Implicit Differentiation

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