Question

The (17291728)^th derivative of f(x) = (x^2 + 1)tan^-1 (x) at x = 0 is

(a) 0

(b) 1729

(c) 1728

(d) ∞

I got this question in class test.

This interesting question is from Leibniz Rule in portion Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (a) 0

The best I can explain: Expanding the tan^-1 (x) function into Taylor series we have

tan^-1(x)=\(\frac{x}{1}-\frac{x^3}{3}+\frac{x^5}{5}-…\infty\)

Rewrite the function as

f(x)=(x^2+1)\((\frac{x}{1}-\frac{x^3}{3}+\frac{x^5}{5}-…\infty)\)

Now observe that the derivative in question is even. Hence, even terms are the only ones to contribute to the derivative at x = 0

Also note that there are no even powered terms in the function. One can conclude that the (17291728)^th derivative at x = 0 is 0.