By Mejlbro L.
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Additional info for Calculus 1c-2, Examples of Elementary Functions
Analysis of a function. D. Diﬀerentiate and start with the second question. I. When x = 0, we get by a diﬀerentiation that f (x) = − 1 1 1+ 2 x − 1 x2 = d 1 Arctan x. = 1 + x2 dx When we integrate again we conclude that Arccot 1 = x Arctan x + c− Arctan x + c+ If x = 1 we have Arccot 1 = for x < 0, for x > 0. π π = Arctan 1 + c+ = + c+ , dvs. c+ = 0. 4 4 For x = −1 we have Arccot(−1) = 3π π = Arctan(−1) + c− = + c− , dvs. c− = π. 4 4 Finally we conclude that Arccot 1 = x Arctan x + π, Arctan x, for x < 0, for x > 0.
1) Apply the theorem of diﬀerentiation of an inverse function. 2) Follow the guideline above. 3) Modify the method from (2). 4) Note that y = Arctan x is the inverse of x = tan y and then use (2). It is also possible alternatively to prove the claim directly. DIVERSE - INNOVATIVE - INTERNATIONAL Please click the advert Are you considering a European business degree? Copenhagen Business School is one of the largest business schools in Northern Europe with more than 15,000 students from Europe, North America, Australia and Asia.
Com 35 Calculus Analyse c1- 2 Some Functions known from High School 2) A replacement of x by xα , x > 0, α > 0 in the estimate above gives α ln x = ln (xα ) < xα , hence by a division by α > 0, ln x < 1 α x , α x > 0, 3) When we put α = ln x < α > 0. β , β > 0 in (2), we get 2 2 β/2 x , β hence by insertion, 2 xβ/2 2 ln x < = · x−β/2 , β x β xβ β x > 0. Now β > 0 and xβ/2 → +∞ for x → +∞, so we conclude that ln x →0 xβ for x → +∞ for every β > 0. Remark. Putting x = et , t ∈ R+ , we get from (3) that ln x 2 β t < exp − t = xβ eβt β 2 →0 for t → +∞.
Calculus 1c-2, Examples of Elementary Functions by Mejlbro L.