Tony Spark 194 Questions 0 Answers 0 Best Answers 1k Points View Profile Tony SparkEnlightened Asked: April 25, 20202020-04-25T15:50:47+00:00 2020-04-25T15:50:47+00:00In: Calculus questions and answers The nth derivative of f (x) = 1/1 – x^2 is (A) 1/(1 – x^2)^n + 1 (B) x ^n/(1 – x^2)^n + 1 (C) … 1 Answer Voted Oldest Recent Anonymous Added an answer on April 25, 2020 at 3:50 pm f(x) = 1/1 – x^2 = 1/(1 + x)(1 + x) using partial fraction I can write it 1/(1 + x)(1 – x) = A/(1 + x) + B/(1 – x) 1 = A(1 -x) + B(1 + x) putting x = 1 we get B = 1/2 putting x = 1 we get A = 1/2 so 1/(x + 1) (1 – x) = 1/2[1(1/(1 + x) + 1/(1 + x)] f(x) = 1/2[1/(1 -x) + 1/(1 + x)] f^n(x) = 1/2[d^n/d x^n 1/d x + d^n/d x^n 1/(1 + x)] using formula d^n/d x^n [1/ax + b] = (-1)^n/a^n/(ax + b)^n + 1 rn putting the value f^n(x) = 1/2[(-1)^n n!(-1)^n/(-x + 1)^n + 1 + (-1)^n n! (1)^n/(x + 1)^n + 1] n!/2[(-1)^2 n/(1-x)^n + 1 + (-1)^n/(x + 1)^n + 1] = n!/2 [1/(1 – x)^n + 1 + (-1)^n/(1 + x)^n + 1 + (-1)^n/(1 + x)^n + 1] so option (E) is correct 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Answer Anonymously Save my name, email, and website in this browser for the next time I comment.

f(x) = 1/1 – x^2 = 1/(1 + x)(1 + x) using partial fraction I can write it 1/(1 + x)(1 – x) = A/(1 + x) + B/(1 – x) 1 = A(1 -x) + B(1 + x) putting x = 1 we get B = 1/2 putting x = 1 we get A = 1/2 so 1/(x + 1) (1 – x) = 1/2[1(1/(1 + x) + 1/(1 + x)] f(x) = 1/2[1/(1 -x) + 1/(1 + x)] f^n(x) = 1/2[d^n/d x^n 1/d x + d^n/d x^n 1/(1 + x)] using formula d^n/d x^n [1/ax + b] = (-1)^n/a^n/(ax + b)^n + 1 rn putting the value f^n(x) = 1/2[(-1)^n n!(-1)^n/(-x + 1)^n + 1 + (-1)^n n! (1)^n/(x + 1)^n + 1] n!/2[(-1)^2 n/(1-x)^n + 1 + (-1)^n/(x + 1)^n + 1] = n!/2 [1/(1 – x)^n + 1 + (-1)^n/(1 + x)^n + 1 + (-1)^n/(1 + x)^n + 1] so option (E) is correct