Question

Let f(x) = (x^2 + x + 1)sinh(x) the (1097)^th derivative at x = 0 is

(a) 1097

(b) 1096

(c) 0

(d) 1202313

The question was posed to me by my college director while I was bunking the class.

I would like to ask this question from Leibniz Rule in section Differential Calculus of Engineering Mathematics

Answer 1

The correct choice is (d) 1202313

For explanation I would say: Expanding sinh(x) into a taylor series we have

sinh(x)=\(x+\frac{x^3}{3!}+\frac{x^5}{5!}…\infty\)

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

On multiplication we get two series with odd exponents and one series with even exponent. The series with odd exponents are the only ones to contribute to the derivative at x=0

Hence it is enough to compute the   derivative at   for the following function

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

Taking the 1097^thderivative of this function, we have

f^(1097)(x)=\((1097)!(\frac{1}{(1097)!}+\frac{1}{(1095)!})+(1099\times 1098…4\times 3)x^2(\frac{1}{(1097)!}+\frac{1}{(1097)!})+…\infty\)

Substituting x=0 we have

f^(1097)(x)=\((1097)!(\frac{1}{(1097)!}+\frac{1}{(1095)!})\)

=(1+1097*1096)=(1+1202312)=1202313