Question

Let f(x) = e^x sin(x^2) ⁄ x Then the value of the fifth derivative at x = 0 is given by

(a) 25

(b) 21

(c) 0

(d) 5

The question was posed to me during an online interview.

Question is taken from Leibniz Rule topic in section Differential Calculus of Engineering Mathematics

Answer 1

Right option is (b) 21

The best I can explain: First expanding ^sin(x^2) ⁄x into a Taylor series we have

sin(x^2)=\(\frac{x^2}{1!}-\frac{x^6}{3!}+\frac{x^{10}}{5!}….\infty\)

\(\frac{sin(x^2)}{x}=\frac{x}{1!}-\frac{x^5}{3!}+\frac{x^9}{5!}….\infty\)

Now applying the Leibniz rule up to the fifth derivative we have

\(((e^x)(\frac{sin(x^2)}{x})^{(5)} = c_0^5e^x(\frac{x}{1!}-\frac{x^5}{3!}+\frac{x^9}{5!}…\infty)\)

\(+c_1^5e^x(\frac{1}{1!}-\frac{5x^4}{3!}+\frac{9x^8}{5!}…\infty)+…..+c_5^5e^x(\frac{5!}{3!}+\frac{(9.8.7.6.5)x^4}{5!}…\infty)\)

Now substituting x=0 we get

\(((e^x)(\frac{sin(x^2)}{x}))^{(5)}=c_1^5(1)+\frac{5!}{3!}\)

= 1 + 20 = 21.