m2121t3rev-sol

m2121t3rev-sol - MATH/2121 CALCULUS FOR THE LIFE SCIENCES I...

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MATH/2121, CALCULUS FOR THE LIFE SCIENCES I TEST 3 REVIEW PROBLEMS SOLUTIONS (1) Compute the composites f ( g ( x )) and g ( f ( x )) for the following pairs of func- tions f , g . (a) f ( x ) = x , g ( x ) = sin( x + 1). f ( g ( x )) = p sin( x + 1) , g ( f ( x )) = sin( x + 1) . (b) f ( x ) = x 2 + x - 2, g ( x ) = x 3 . f ( g ( x )) = ( x 3 ) 2 + x 3 - 2 = x 6 + x 3 - 2 , g ( f ( x )) = ( x 2 + x - 2) 3 . (c) f ( x ) = e 3 x , g ( x ) = cos( x 2 ). f ( g ( x )) = e 3 cos( x 2 ) , g ( f ( x )) = cos(( e 3 x ) 2 ) = cos( e 6 x ) . (d) f ( x ) = 1 / ( x 2 + 1), g ( x ) = ln( x + 2). f ( g ( x )) = 1 (ln( x + 2)) 2 + 1 , g ( f ( x )) = ln 1 x 2 + 1 + 2 = ln 2 x 2 + 3 x 2 + 1 . (2) Write each of the following as a composition f ( g ( x )) of two functions f , g . (a) y = (6 - x 2 ) 3 / 2 . y = f ( g ( x )) for f ( x ) = x 3 / 2 and g ( x ) = 6 - x 2 . (b) y = sin( x 2 + x ). y = f ( g ( x )) for f ( x ) = sin( x ) and g ( x ) = x 2 + x . (c) y = (ln( x 2 + 1)) 3 . y = f ( g ( x )) for f ( x ) = x 3 and g ( x ) = ln( x 2 + 1). (d) y = e 3 x . y = f ( g ( x )) for f ( x ) = e x and g ( x ) = 3 x . (e) y = 1 / ( x 3 + x + 4). y = f ( g ( x )) for f ( x ) = 1 /x = x - 1 and g ( x ) = x 3 + x + 4. (3) Differentiate the following functions. Simplify your answers as appropriate. 1

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(a) f ( t ) = 1 2 t 6 - 3 t 4 + t has derivative f 0 ( t ) = 3 t 5 - 12 t 3 + 1 . (b) y = 3 x 1 / 2 + 4 x 1 / 3 + 5 + 6 x - 1 / 3 + 7 x - 1 / 2 has derivative dy dx = 3 2 x - 1 / 2 + 4 3 x - 2 / 3 - 2 x - 4 / 3 - 7 2 x - 3 / 2 . (c) g ( x ) = 3 x - 1 2 x + 1 has derivative g 0 ( x ) = (2 x + 1) × d dx (3 x - 1) - (3 x - 1) × d dx (2 x + 1) (2 x + 1) 2 = (2 x + 1) × 3 - (3 x - 1) × 2 (2 x + 1) 2 = 6 x + 3 - 6 x + 2 (2 x + 1) 2 = 5 (2 x + 1) 2 . (d) y = t - 1 t has derivative dy dt = d dt ( t 1 / 2 - t - 1 / 2 ) = 1 2 t 1 / 2 - 1 - - 1 2 t - 1 / 2 - 1 = 1 2 t - 1 / 2 + 1 2 t - 3 / 2 = 1 2 t + 1 2 t 3 . (e) f ( x ) = 30 has derivative f 0 ( x ) = 0 . (The derivative of a constant is zero!) (f) y = x 2 + 1 has derivative dy dx = d dx ( ( x 2 + 1) 1 / 2 ) = 1 2 ( x 2 + 1) 1 / 2 - 1 × d dx ( x 2 + 1) = 1 2 ( x 2 + 1) - 1 / 2 × 2 x = x ( x 2 + 1) - 1 / 2 = x x 2 + 1 . 2
(g) y = 2 x x 2 + 1 has derivative dy dx = 2 d dx ( x ( x 2 + 1) 1 / 2 ) = 2 dx dx × ( x 2 + 1) 1 / 2 + x × d dx ( x 2 + 1) 1 / 2 = 2 ( (1)( x 2 + 1) 1 / 2 + x ( x ( x 2 + 1) - 1 / 2 )) (using (f)) = 2( x 2 + 1) 1 / 2 + 2 x 2 ( x 2 + 1) - 1 / 2 = 2( x 2 + 1) 1 / 2 + 2 x 2 ( x 2 + 1) 1 / 2 = 2 ( ( x 2 + 1) 1 / 2 ) 2 + 2 x 2 ( x 2 + 1) 1 / 2 = 2( x 2 + 1) + 2 x 2 ( x 2 + 1) 1 / 2 = 2 x 2 + 2 + 2 x 2 ( x 2 + 1) 1 / 2 = 4 x 2 + 2 ( x 2 + 1) 1 / 2 = 4 x 2 + 2 x 2 + 1 . (h) h ( t ) = t 2 cos( t ) + 1 has derivative h 0 ( t ) = d dt ( t 2 ) × cos( t ) + t 2 × d dt (cos( t )) + 0 = 2 t cos( t ) + t 2 ( - sin( t )) = 2 t cos( t ) - t 2 sin( t ) . (i) y = x + 1 3 x 4 has derivative dy dx = d dx ( x 1 / 2 + x - 4 / 3 ) = 1 2 x - 1 / 2 - 4 3 x - 7 / 3 = 1 2 x - 4 3 3 x 7 (j) f ( x ) = sin( x ) has derivative f 0 ( x ) = d dx ( sin( x 1 / 2 ) ) = sin 0 ( x 1 / 2 ) × d dx ( x 1 / 2 ) = cos( x 1 / 2 ) × 1 2 x - 1 / 2 = cos( x 1 / 2 ) 2 x 1 / 2 = cos( x ) 2 x . 3

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(k) f ( x ) = 10 e 2 x +4 has derivative f 0 ( x ) = 10 d dx ( e 2 x +4 ) = 10 e 2 x +4 × d dx (2 x + 4) = 10 e 2 x +4 × 2 = 20 e 2 x +4 . (l) f ( x ) = sin( e x 2 ) has derivative f 0 ( x ) = cos( e x 2 ) × d dx ( e x 2 ) = cos( e x 2 ) × e x 2 × d dx ( x 2 ) = 2 xe x 2 cos( e x 2 ) . (m) f ( x ) = e cos( x ) has derivative f 0 ( x ) = e cos( x ) × d dx (cos( x )) = - sin( x ) e cos( x ) . (n) f ( x ) = x 2 e - 3 x has derivative f 0 ( x ) = d dx ( x 2 ) × e - 3 x + x 2 × d dx ( e - 3 x ) = 2 xe - 3 x + x 2 e - 3 x × d dx ( - 3 x ) = 2 xe - 3 x + x 2 e - 3 x ( - 3) = 2 xe - 3 x - 3 x 2 e - 3 x = (2 x - 3 x 2 ) e - 3 x . (o) f ( x ) = xe x 3 has derivative f 0 ( x ) = d dx ( x ) × e x 3 + x × d dx ( e x 3 ) = 1 e x 3 + xe x 3 × d dx ( x 3 ) = e x 3 + xe x 3 3 x 2 = e x 3 + 3 x 3 e x 3 = (1 + 3 x 3 ) e x 3 . (p) F ( t ) = t cos( t ) has derivative F 0 ( t ) = cos( t ) × d dt ( t ) - t × cos 0 ( t ) (cos( t )) 2 = cos( t ) × 1 - t × ( - sin( t )) cos 2 ( t ) = cos( t ) + t sin( t ) cos 2 ( t ) .
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