England_MAT_265_ONLINE_B_Summer_2019.jlalonzo.Section_3.3.pdf

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Jesse Alonzo England MAT 265 ONLINE B Summer 2019 Assignment Section 3.3 due 07/28/2019 at 11:59pm MST 1. (1 point) Let f ( x ) = - 5ln ( 4 x ) f 0 ( x ) = f 0 ( 5 ) = Solution: Using the chain rule gives f 0 ( x ) = - 5 4 x d dx ( 4 x ) = - 5 4 x ( 4 ) = - 5 x Evaluating the derivative at x = 5 gives: f 0 ( 5 ) = - 1 Answer(s) submitted: -5/x -1 (correct) Correct Answers: -(5/x) -1 2. (1 point) Let f ( x ) = ( ln x ) 2 f 0 ( x ) = f 0 ( e 4 ) = Solution: SOLUTION f 0 ( x ) = 2 ( ln x ) 1 d dx ( ln x ) = 2 ( ln x ) 1 · 1 x f 0 ( e 4 ) = 2 ( ln ( e 4 ) 1 1 e 4 = 2 · 4 1 · 1 e 4 = 8 e 4 Answer(s) submitted: 2ln(x) / x 0.146525111 (correct) Correct Answers: 2 / x *(ln(x))ˆ(2 - 1) 0.14652511120885 3. (1 point) Let f ( x ) = 2 x 4 ln x f 0 ( x ) = f 0 ( e 4 ) = Solution: Using the product rule gives f 0 ( x ) = 8 x 3 ln x + 2 x 4 1 x = 8 x 3 ln x + 2 x 3 = 2 x 3 ( 4ln ( x )+ 1 ) Evaluating the derivative at x = e 4 and remembering that ln ( e p ) = p , gives: f 0 ( e 4 ) = 2 ( e 4 ) 3 ( 4ln ( e 4 ) + 1 ) = 2 e 12 · 17 = 34 e 12 5 . 53366 × 10 6 Answer(s) submitted: 2(4xˆ3ln(x)+xˆ3) 200.8553692 (score 0.5) Correct Answers: 2*xˆ3*[4*ln(x)+1] 34*eˆ12 4. (1 point) Let y = ln (( x + 3 ) 10 ) dy dx = Solution: We first rewrite the function using properties of logarithms: y = ln (( x + 3 ) 10 ) = 10ln ( x + 3 ) The derivative is then dy dx = 10 x + 3 Answer(s) submitted: 10/(x+3) (correct) Correct Answers: 10/(x+3) 5. (1 point) Let y = ln ( 6 t 2 + 11 t + 4 ) dy dt = Solution: The chain rule yields: dy dt = 1 6 t 2 + 11 t + 4 d dt ( 6 t 2 + 11 t + 4 ) = 12 t + 11 6 t 2 + 11 t + 4 Answer(s) submitted: 12t / ( 6tˆ2 + 11y + 4 ) (incorrect) Correct Answers: (12*t+11)/(6*tˆ2+11*t+4) 1
6. (1 point) Let f ( x ) = 3 ln ( x 2 + 4 ) . f 0 ( 1 ) = Solution: We first rewrite the function as f ( x ) = 3 ( ln ( x 2 + 4 ) ) - 1 Using the chain rule we get: f 0 ( x ) = - 3 ( ln ( x 2 + 4 ) ) - 2 d dx ( ln ( x 2 + 4 ) ) = - 3 ( ln ( x 2 + 4 )) 2 2 x x 2 + 4 Evaluting the derivative at x = 1 yields f 0 ( 1 ) = - 3 ( ln ( 5 )) 2 2 5 = - 6 5 1 ( ln ( 5 )) 2 ≈ - 0 . 463269 Answer(s) submitted: -0.72134752044 (incorrect) Correct Answers: -0.463269 7. (1 point) Evaluate d dx 4 q ln ( 9 - x 2 ) at x = 1. Answer: Solution: d dx 4 q ln ( 9 - x 2 ) = d dx ( ln ( 9 - x 2 ) ) 1 / 4 = 1 4 ( ln ( 9 - x 2 ) ) - 3 4 d dx ln ( 9 - x 2 ) = 1 4 ( ln ( 9 - x 2 ) ) - 3 4 - 2 x 9 - x 2 Evaluating the derivative at x = 1 we get - 1 16 ( ln ( 8 )) - 3 4 ≈ - 0 . 0360928 Answer(s) submitted: -1 / (16ln(8)ˆ(3/4)) (correct) Correct Answers: 2*[ln(8)]ˆ(-1+1/4)/-32 8. (1 point) Let f ( x ) = 7 x log 5 ( x ) f 0 ( x ) = help (logarithms) Solution: Recall that ( a x ) 0 = ln ( a ) a x and ( log a ( x )) 0 = 1

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