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# S1 - Math 128 Solutions to Assignment 1 Koray...

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Math 128: Solutions to Assignment 1 Koray Karabina ([email protected]) 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 ) 0 + e sin x (sin x ) 0 = (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) 0 ( x 2 + 5) = 3 x 2 + 2 · y · y 0 2 x x 2 + 5 = 3 x 2 + 2 · p ln ( x 2 + 5) - x 3 · y 0 Solving for y 0 and simplifying the equation yields: y 0 = - 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 0 ( x ) = 1 / (1 + x 2 ), g 0 ( x ) = 2 x , and using the chain rule we have: y 0 = f 0 ( g ( x )) g 0 ( x ) = 1 1 + ( x 2 + 1) 2 · 2 x

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Solution 2 (Implicit differentiation): y = tan - 1 ( x 2 + 1) implies tan y = x 2 + 1. Differentiating this last equation with respect to x , we have: dy dx [(tan y )] = dy dx [( x 2 + 1)] (sec 2 y ) · y 0 = 2 x y 0 = 2 x · (cos 2 y ) (1)
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