Question

If nth derivative of y = \(\frac{x}{(x+1)(x+2)}\) is yn = a(-1)^n+! n! (x+1)^-n-1 + b(-1)^nn!(x+2)^-n-1 then find the value of a and b.

(a) -1, -2

(b) 2, 1

(c) 1, 2

(d) -2, -1

The question was asked during an online exam.

My enquiry is from The nth Derivative of Some Elementary Functions topic in section Differential Calculus of Engineering Mathematics

Answer 1

The correct answer is (c) 1, 2

To explain I would say: y = \(\frac{x}{(x+1)(x+2)}\)

y = \(\frac{(-1)}{(x+1)} + \frac{2}{(x+2)}\)

y = (-1)(-1)^nn!(x+1)^-n-1 + 2(-1)^nn!(x+2)^-n-1

y = (-1)^n+1n!(x+1)^-n-1 + 2(-1)^nn!(x+2)^-n-1

Hece, a=1 and b=2.