Finding the 6th Derivative of a function of $\sin(x)$

Question

I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:

$$f(x) = \frac{1}{1+\sin(x^2)}$$

The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $\sin(x)$, but I get:

$$\frac{1}{1+\sin(x^2)} = \frac{1}{1+x^2-\frac{x^6}{6!}+\frac{x^{10}}{5!}}$$

I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.

Answer 1

You're on the right track.

Continuing, we see that

$$\begin{align} \frac{1}{1+\sin(x^2)}&=\frac{1}{1+\left(x^2-\frac16x^6+O\left(x^{10}\right)\right)}\\\\ &=1-\left(x^2-\frac16x^6+O\left(x^{10}\right)\right)+\left(x^2+O\left(x^6\right)\right)^2-\left(x^2+O\left(x^6\right)\right)^3+O(x^8)\\\\ &=1-x^2+x^4-\frac56x^6+O(x^8) \end{align}$$

Therefore, the sixth derivative is $-(5/6)6!=-600$.

Answer 2

Use the geometric series i.e. $$\frac{1}{1 + \sin(x^2)} = 1 + \Big( -x^2 + \frac{x^6}{3!} - \frac{x^{10}}{5!} + ... \Big) + \Big( - x^2 + \frac{x^6}{3!} - \frac{x^{10}}{5!} + ... \Big)^2 + ...$$

Answer 3

You properly wrote $$\frac{1}{1+\sin(x^2)} = \frac{1}{1+x^2-\frac{x^6}{6!}+\frac{x^{10}}{5!}+O\left(x^{12}\right)}$$ Now perform the long division (by increasing powers of $x$) to get $$\frac{1}{1+\sin(x^2)} = 1-x^2+x^4-\frac{5 x^6}{6}+\frac{2 x^8}{3}-\frac{61 x^{10}}{120}+O\left(x^{12}\right)$$ Therefore

• the sixth derivative is $-\frac 5 6\times 6!=-600$
• the eighth derivative is $\frac 2 3\times 8!=26880$
• the tenth derivative is $-\frac {61}{120}\times 10!=-1844640$