3

If $\displaystyle y=\frac{\sin(x^2)-x^2}{x^3}$.

Then value of $\displaystyle \frac{d^7y}{dx^7}\bigg|_{x=0}=$

What i try

$$\sin x=\sum^{\infty}_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$$

Replace $x\rightarrow x^2$

Then $$\sin (x^2)=\sum^{\infty}_{n=0}(-1)^n\frac{x^{4n+2}}{(2n+1)!}$$

$$\frac{\sin(x^2)-x^2}{x^3}=\sum^{\infty}_{n=1}(-1)^n\frac{x^{4n-1}}{(2n+1)!}$$

How do i find its $7$ th derivative. Although i have calculate $1$ st or $2$ nd derivative. But $7$ th derivative is very conplex.

Please help me How to solve it. Thanks

Community
  • 1
jacky
  • 5,194

3 Answers3

5

You have done most of the work ,since $$ \frac{\sin(x^2)-x^2}{x^3}=\sum^{\infty}_{n=1}(-1)^n\frac{x^{4n-1}}{(2n+1)!}=-\frac{x^3}{3!}+\frac{x^7}{5!}-•••$$

the n-th term of Taylor's expansion of a function $f(x)$ is $$\frac{f^n(0)}{n!}$$ Therefore $\frac{f^n(0)}{n!} =(7!)/(5!)=42$

Nimu Basak
  • 1,120
3

You have: $$ y(x)=-\frac{x^3}{3!}+\frac{x^7}{5!}-\frac{x^{11}}{7!}+\cdots, $$ so that $$ y^{(7)}(0)=\frac{7!}{5!}=42. $$

user
  • 26,272
1

Hint. For any power $y=x^\alpha,$ where $\alpha$ is a nonnegative integer, the $n$th derivative is given by $$\frac{\alpha!}{(\alpha-n)!}x^{\alpha-n},$$ where we must ensure that $n\le\alpha,$ or better still just declaring that the coefficient of the $n$th derivative vanish whenever $n>\alpha.$

This can easily be generalised, and the caveat above dispensed with, by using the $\Gamma$-function, but you don't need that here.

Allawonder
  • 13,327