Question

Let f(x)=\(\frac{e^x \times sin(x)}{x}\) and let the n^th derivative at x = 0 be given by y^(n) Then the value of the expression for y^(n) is given by

(a) \(\frac{\pi n}{4}\)

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

(c) πn

(d) \(\frac{\pi}{2n}\)

This question was addressed to me in class test.

The above asked question is from Leibniz Rule in division Differential Calculus of Engineering Mathematics

Answer 1

Correct answer is (b) \(\sum_{i=0}^{i<=n}\frac{(-1)^i c_{2i}^n}{2i+1}\)

To explain I would say: Expanding sin(x)/x into Taylor series we have

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

Now Taking the n^th derivative of function using Leibniz rule we have

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

Now substituting x=0 we have

\([(e^x(\frac{sin(x)}{x}))^{(n)}]_{x=0} = c_0^n(\frac{1}{1!})+c_2^n(\frac{-2!}{3!})+c_4n(\frac{4!}{5!})…\infty\)

\([(e^x(\frac{sin(x)}{x}))^{(n)}]_{x=0}=\sum_{i=0}^{i<=n}\frac{(-1)^i c_{2i}^n}{2i+1}\)

Hence, Option \(\sum_{i=0}^{i<=n}\frac{(-1)^i c_{2i}^n}{2i+1}\) is the right answer.