S1 - Math 128: Solutions to Assignment 1 Koray Karabina

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Unformatted text preview: Math 128: Solutions to Assignment 1 Koray Karabina (kkarabin@uwaterloo.ca) 1. Find dy/dx for the following (show your work): (a) [5 points] y = x 2 sin( x ) + e x tan( x ) (b) [6 points] y = sin 3 ( x ) + e sin( x ) Hint: Write e sin( x ) = f ( g ( x )), where f ( x ) = e x and g ( x ) = sin( x ); and use the Chain rule. (c) [5 points] ln( x 2 + 5) = x 3 + y 2 (d) [5 points] y = tan- 1 ( x 2 + 1) (e) [5 points] y = ln( x 2 + x + 1) (f) [6 points] y = x 3 sin( x ) Hnt: Use logarithmic differentiation. Solution: (a) Using the product rule together with differentiation rules for polynomials, trigonomet- ric/exponential functions we find: dy dx = 2 x sin x + x 2 cos x + e x tan x + e x sec 2 x (b) Using the chain rule we compute: dy dx = 3(sin 2 x )(sin x ) + e sin x (sin x ) = (3 sin 2 x + e sin x ) cos x (c) We differentiate both sides with respect to x : dy dx [ln( x 2 + 5)] = dy dx [( x 3 + y 2 )] ( x 2 + 5) ( x 2 + 5) = 3 x 2 + 2 y y 2 x x 2 + 5 = 3 x 2 + 2 p ln ( x 2 + 5)- x 3 y Solving for y and simplifying the equation yields: y =- 3 x 4- 15 x 2 + 2 x 2( x 2 + 5) p ln ( x 2 + 5)- x 3 (d) There are (at least) two solutions to this question: Solution 1 (Chain rule): Let f ( x ) = tan- 1 ( x ) and g ( x ) = x 2 + 1. Then y = f ( g ( x )), f ( x ) = 1 / (1 + x 2 ), g ( x ) = 2 x , and using the chain rule we have: y = f ( g ( x )) g ( x ) = 1 1 + ( x 2 + 1) 2 2 x Solution 2 (Implicit differentiation):...
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This note was uploaded on 01/26/2012 for the course ECON 401 taught by Professor Burbidge,john during the Fall '08 term at Waterloo.

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S1 - Math 128: Solutions to Assignment 1 Koray Karabina

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