Question

The fourth derivative of f(x) = sin(x)sinh(x) ⁄ x at x = 0 is given by

(a) 0

(b) ^π⁄2

(c) 45

(d) 4

The question was asked in unit test.

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

Answer 1

Correct answer is (a) 0

Best explanation: First convert the function sinh(x)⁄x into its Taylor series expansion

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

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

Now pick up the whole function \(((sin(x))(\frac{sinh(x)}{x}))\) and apply Leibniz rule up to the fourth derivative we have

\(((sin(x))(\frac{sinh(x)}{x}))^{(4)}=c_0^4sin(x)(\frac{1}{1!}+\frac{x^2}{3!}+\frac{x^4}{5!}…..\infty)\) \(-c_1^4cos(x)(\frac{2x}{3!}+\frac{4x^3}{5!}…..\infty)+…..+c_4^4sin(x)(\frac{4!}{5!}+\frac{(7.6.5.4)x^3}{7!}…..\infty)\)

Substituting x=0 we have

\(((sin(x))(\frac{sinh(x)}{x}))^{(4)}\) = 0