Question

Let f(x)=\(\frac{ln(1-x)}{e^x}\). Find the third derivative at x = 0.

(a) 4

(b) ^1⁄3

(c) Undefined

(d) ^1⁄4

This question was posed to me by my school teacher while I was bunking the class.

Asked question is from The nth Derivative of Some Elementary Functions topic in portion Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (b) ^1⁄3

Easiest explanation: Again the key here is to expand the given function into appropriate Taylor series.

Rewriting the function as f(x) = e^-x(ln(1 – x)) and then expanding into Taylor series we have

\(f(x)=(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}…\infty) \times (\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}…\infty)\)

Now the question asks us to find the third derivative at x = 0. Thus, it is enough for us to find the coefficient of the x^3 term in the infinite polynomial product above

The third degree terms can be grouped apart as follows

= ^x^3⁄3 – ^x^3⁄2 + ^x^3⁄2

Hence the third derivative at x = 0 is simply the coefficient of the third degree term, which is

coefficient(^x^3⁄3) = ^1⁄3.