University:
University of Michigan
Course:
MATH 115 | Calculus I
Academic year:

2024
Views:

302

Pages:

3
Author:

Luciano Kirby

Question 1 How to find first derivative of function y = xln (x) by limit definition, that is using this formula y’ = lim (h→0) [f(x+h) - f(x)] / h Answer 1 lim (h→0) [(x+h)ln(x+h) - xlnx] / h = lim(h→0) [x(ln(x+h)-ln(x)) + hln(x+h)] / h = lim(h→0) x [ln(1 + (h/x)) / h] + lim(h→0) ln(x+h) = x * lim(h→0) [ln(1 + (h/x)) / (h/x) * (1/x)] + lim(h→0) ln(x+h) = x * (1/x) + ln x = 1 + ln x where xln * lim(h→0) [ln(1 + (h/x)) / (h/x) ] = 1 follows from the well known limit: lim(h→0) (1 + (h/x))^(x/h) = e Answer 2 We have that lim(h→0) [(x+h)log(x+h) - xlog x] / h = lim(h→0) [x(log (x+h) - log x) + hlog(x+h)] / h = lim(h→0) x * (log (x+h) - log x) / h + lim(h→0) log(x+h) = lim(h→0) x * (1 / x) * log(1 + (h/x)) / (h/x) + log x = x * (1 / x) + log x = 1 + log x indeed lim(h→0) (log (x+h) - log x) / h = (1/x) lim(y→0) log(1 + y) / y = 1/x indeed by y = h/x → 0 lim(y→0) log(1+y)/y = 1 Answer 3 Continuing with the suggestion in my comment, you might do lim(h→0) [(x+h) ln(x+h)-xlnx] / h = lim(h→0) [(x+h) ln(x+h) - xln(x+h) + xln(x+h)- xlnx] / h = lim(h→0) ln(x+h) * [ (x+h) - x ] / h + lim(h→0) x * [ ln(x+h) - ln x ] / h = lim(h→0) ln(x+h) + x * lim(h→0) ln(1 + (h/x)) / h x+h-x = h, so the first limit is 1. Question 2 How to prove the identity sin 2x + sin 2y = 2sin(x + y)cos(x - y) Answer 2sin (x + y)cos (x - y) = 2(sin x cos y + cos x sin y) (cos x cos y + sin x sin y) = 2(sin x cos x(cos²y + sin²y) + 2 sin y cos y (cos²x + sin²x)) = 2 sin x cos x + 2 sin y cos y = sin 2x + sin 2y Question 3 I calculated sin 75° as (1 / 2√2) + (√3 / 2√2) but the answer is (√2 + √6) / 4. What went wrong? I calculated the exact value of sin 75° as follows: sin 75°=sin (30° +45°) = sin 30° cos 45° + cos 30° sin 45° = 1/2 * 1/√2 + √3/2 * 1/√2 = 1/√2 + √3/ 2√2 The actual answer is (√2 + √6) / 4 Answer They're the same value. Multiply the numerator and denominator of your answer by √2 to see why. (1 + √3) / (2√2) * (√2 / √2) = (√2 + √6) / 4 You can also use sin 45° = cos 45° = (√2 / 2) (rationalizing 1/√2) to get the answer more easily. Question 4 Identities similar to arctan (x) + arctan (1/x) = π/2 The arctan (x) + arctan (1/x) = π/2 identity can be solved by taking the derivative of the left hand side, showing it is 0, and then plugging in, say, x=1 to get its constant value π/2 Are there any other (nontrivial) identities which can be solved similarly? Answer 1 Answer provided below arccos(x) + arcsin(x) = π/2 Answer 2 If we know sin'(x) = cos (x) and cos'(x) = -sin (x), then we can compute d/dx [sin²(x) + cos²(x)] = 0, to conclude sin²(x) + cos²(x) is constant. Then plug in one value to get sin²(x) + cos²(x) = 1 Question 5 Double Angle Formulas: Finding tan 2θ I am trying to find tan 2θ where sin θ = 5/13 and θ is in Quadrant One. According to my textbook, tan 2θ = 120/119, but I get -10/13 instead. The Identity I am using: tan 2θ = (2 tan θ) / (1 - tan²θ) Answer tan θ = 5/12 tan 2θ = (2 * (5/12)) / (1 - (5/12)²) = (10/12) / (1 - (25/144)) = (10/12) / ((144 - 25)/144) = (10/12) * (144/119) = (10 * 12) / 119 = 120/119 Therefore, tan 2θ = 120/119