Question

Let g(x) = ln(x) ⁄ x – 1 Then the hundredth derivative at x = 1 is?

(a) ^100!⁄101

(b) ^99!⁄101

(c) ^101⁄100!

(d) ^1⁄99!

This question was addressed to me during an online interview.

This interesting question is from The nth Derivative of Some Elementary Functions in division Differential Calculus of Engineering Mathematics

Answer 1

Correct option is (a) ^100!⁄101

To elaborate: The key here is to again expand the numerator as a Taylor series centered at x = 1

Hence we have the Taylor series as

ln(x)=\(\frac{(x-1)}{1}-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}…..\infty\)

Hence our function  g(x) transforms into

g(x)=\(1-\frac{(x-1)}{2}-\frac{(x-1)^2}{3}+\frac{(x-1)^3}{4}…..\infty\)

This now simplifies into polynomial differentiation and the hundredth derivative can be written as

g^(100)(x)=\(\frac{100!}{101}-\frac{101!\times(x-1)}{102}…..\infty\)

Substituting x = 1 yields

^100!⁄101.