Exam 1 solutions

# Exam 1 solutions - 18.01 Solutions to Exam 1 Problem 1(15...

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18.01 Solutions to Exam 1 Problem 1 (15 points) Use the definition of the derivative as a limit of difference quotients to 1 compute the derivative of y = x + x for all points x > 0. Show all work. 1 Solution to Problem 1 Denote by f ( x ) the function x + x . By definition, the derivative of f ( x ) at x = a is, f ( a ) = lim f ( a + h ) f ( a ) . h 0 h The increment f ( a + h ) f ( a ) equals, 1 1 1 1 ( a + h ) + a + = h + . a + h a a + h a To compute the second term, clear denominators, 1 1 1 a 1 a + h a ( a + h ) h = = = . a + h a a + h a a a + h a ( a + h ) a ( a + h ) Thus the increment f ( a + h ) f ( a ) equals, h . h a ( a + h ) Factoring h from each term, the difference quotient equals, f ( a + h ) f ( a ) = 1 1 . h a ( a + h ) Thus the derivative of f ( x ) at x = a equals, 1 1 1 f ( a ) = lim = 1 = 1 . h 0 1 a ( a + h ) a ( a + 0) a 2 Therefore the derivative function of f ( x ) equals, f ( x ) = 1 1 x 2 . 1

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Problem 2 (10 points) For the function f ( x ) = e x 2 / 2 , compute the first, second and third deriva- tives of f ( x ). Solution to Problem 2 Set u equals x 2 / 2 and set v equals e u . So v equals f ( x ). By the chain rule, dv dv du = . dx du dx u Since v equals e u , dv/du equals ( e u ) = e . Since u equals x 2 / 2, du/dx equals (2 x ) / 2 = x . Thus, back-substituting, f ( x ) = dv = ( e u )( x ) = e x 2 / 2 ( x ) = dx xe x 2 / 2 . For the second derivative, let u and v be as defined above, and set w equals xv . So w equals f ( x ). By the product rule, dw dv = ( x ) v + ( x ) v = v x . dx dx By the last paragraph, dv = xe x 2 / 2 . dx Substituting in, f �� ( x ) = dw = e x 2 / 2 x ( xe x 2 / 2 ) = e x 2 / 2 + x 2 e x 2 / 2 = dx ( x 2 1) e x 2 / 2 . For the third derivative, take u and v as above, and set z equals ( x 2 1) v . So z equals f �� ( x ). By the product rule, dv dz = ( x 2 1) v + ( x 2 1) v = 2 xv + ( x 2 1) . dx dx By the first paragraph, dv = xe x 2 / 2 . dx Substituting in, 2 2 / 2 2 / 2 f ��� ( x ) = dz = 2 xe x 2 / 2 + ( x 1)( xe x 2 / 2 ) = 2 xe x + ( x 3 + x ) e x = dx ( x 3 x ) e x . + 3 2 2 / 2
Extra credit (5 points) Only attempt this after you have completed the rest of the exam and checked your answers. For every positive integer n , show that the n th derivative of f ( x ) is of the form f ( n ) ( x ) = p n ( x ) f ( x ), where p n ( x ) is a polynomial. Also, give a rule to compute p n +1 ( x ), given p n ( x ). Solution to extra credit problem The claim, proved by induction on n , is that for every positive integer n , f ( n ) ( x ) equals p n ( x ) where p n ( x ) is a degree n polynomial and, p n +1 ( x ) = xp n ( x ) + p n ( x ) .

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