Question

Let f(x) = e^x sinh(x) / x, let y^(n) denote the n^th derivative of f(x) at x = 0 then the expression for y^(n) is given by

(a) \(\sum_{i=0}^n \frac{c_{2i}^n}{2i+1}\)

(b) \(\sum_{i=0}^n \frac{1}{2i+1}\)

(c) 1

(d) Has no general form

I had been asked this question in my homework.

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

Answer 1

Right option is (a) \(\sum_{i=0}^n \frac{c_{2i}^n}{2i+1}\)

For explanation I would say: The key here is to find a separate Taylor expansion for sinh(x) / x which is

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

\(\frac{sin h(x)}{x}=\frac{x}{1!}+\frac{x^2}{3!}+\frac{x^4}{5!}…\infty\)

Now consider \(((e^x)(\frac{sin h(x)}{x}))\) applying the Leibniz rule for n^th derivative we have

\((e^x(\frac{sin(x)}{x}))^{(n)}=c_0^ne^x(\frac{1}{1!}-\frac{x^2}{3!}+\frac{x^4}{5!}-…\infty)+c_1^ne^x(\frac{2x}{3!}+\frac{4x^3}{5!}…\infty)\)

\(+c_2^n(\frac{2}{3!} + \frac{12x^2}{5!}….\infty)+…\)

Now substituting x=0 yields

\((e^x(\frac{sin(x)}{x}))^{(n)}=\frac{c_0^n}{1} + \frac{c_2^n}{3} + \frac{c_4^n}{5}\)

 \(\sum_{i=0}^n \frac{c_{2i}^n}{2i+1}\)