408Npracticemid2solns[1]

# 408Npracticemid2solns[1] - MATH 408N PRACTICE MIDTERM 2...

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Unformatted text preview: MATH 408N PRACTICE MIDTERM 2 Name: 10/21/2011 Bormashenko TA session: Show your work for all the problems. Good luck! (1) Use the limit definition of the derivative for the following questions. You will get no points for using the rules learned later! (a) [5 pts] Find f (1) if f ( x ) = √ x . Solution: By definition, f (1) = lim h → f (1 + h )- f (1) h = lim h → √ 1 + h- 1 h Now, multiplying top and bottom by the conjugate, then using difference of squares: f (1) = lim h → √ 1 + h- 1 h · √ 1 + h + 1 √ 1 + h + 1 = lim h → ( √ 1 + h ) 2- 1 2 h ( √ 1 + h + 1) = lim h → h h ( √ 1 + h + 1) = lim h → 1 √ 1 + h + 1 Finally, plugging in h = 0 doesn’t result in . Therefore, we plug in, getting that f (1) = 1 √ 1 + 1 = 1 2 Note: You could also use the definition f (1) = lim x → 1 f ( x )- f (1) x- 1 and use a similar trick. (b) [5 pts] Find f ( x ) if f ( x ) = x 2 + x + 1. Solution: By definition, f ( x ) = lim h → f ( x + h )- f ( x ) h = lim h → ( x + h ) 2 + ( x + h ) + 1- ( x 2 + x + 1) h Now, expanding things out, f ( x ) = lim h → x 2 + 2 xh + h 2 + x + h + 1- x 2- x- 1 h = lim h → 2 xh + h 2 + h h = lim h → h (2 x + h + 1) h = lim h → (2 x + h + 1) = 2 x + 1 2 MATH 408N PRACTICE MIDTERM 2 (2) Differentiate the following functions, using whatever methods you think are best (you are now allowed to use all the rules): (a) [5 pts] f ( x ) = x 3 + 3 x + 5 Solution: Using the sum rules and the fact that ( x n ) = nx n- 1 , f ( x ) = 3 x 2 + 3 (b) [5 pts] f ( x ) = arctan ( x 2 ) e x Solution: By the product rule, f ( x ) = ( arctan ( x 2 )) e x + arctan ( x 2 ) ( e x ) Using the chain rule, ( arctan ( x 2 )) = 1 1 + ( x 2 ) 2 · ( x 2 ) = 2 x 1 + x 4 Using straightforward differentiation rules, ( e x ) = e x Now, plugging the expressions for ( e x ) and ( arctan ( x 2 )) back into the f ( x ) formula, f ( x ) = 2 x 1 + x 4 e x + arctan ( x 2 ) e x MATH 408N PRACTICE MIDTERM 2 3 (c) [5 pts] f ( x ) = 2 x sin( x ) ln( x + 1) Solution: Using the quotient rule, f ( x ) = (2 x sin( x )) ln( x + 1)- 2 x sin( x )(ln( x + 1)) (ln( x + 1)) 2 Now, working out the derivative appearing above separately: (2 x sin( x )) = (2 x ) sin( x ) + 2 x (sin( x )) = 2sin( x ) + 2 x cos( x ) ln( x + 1) = 1 x + 1 ( x + 1) = 1 x + 1 where the first one uses product rule, and the second uses chain rule. Plugging those back in, f ( x ) = (2sin( x ) + 2 x cos( x ))ln( x + 1)- 2 x sin( x ) 1 x +1 (ln( x + 1)) 2 (You can expand this out a bit more, but you don’t have to! Careful with the parentheses, though...) (d) [5 pts] f ( x ) = sin( x ) cos( x ) Solution: This one HAS to be done using logarithmic differentiation! Any attempts to use the power rule or the exponent rule will get a wrong answer, because there’s a variable both in the exponent and in the base....
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408Npracticemid2solns[1] - MATH 408N PRACTICE MIDTERM 2...

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