If $f(x)=\cosh(x)+\sinh(x^2)$, what is$f^{(34)}(0)$?
I know that: $h(t)=\sinh(t)=\left(\frac{e^t-e^{-t}}{2}\right)$ and $g(t)=\cosh(t)=\left(\frac{e^t+e^{-t}}{2}\right)$
So, I noticed this:
$g'(t)=\sinh(t)$
$g''(t)=\cosh(t)$
Then: $g^{(34)}(t)=\cosh(t) \rightarrow g^{(34)} (0)= \cosh(0)=1 $
With the same idea, I know that $h^{(34)}(0)=\sinh(0)=0$
The problem that I am having is that I don't know what is the $34^{th}$ derivative of $\sinh(x^2)$, can I use $\sinh(x)$ somehow?
Also, I know that: $\sinh(x)= \sum_{k=0}^{n} \frac{(x^2)^{1+2k}}{(1+2k)!}$
Since$$\sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots,$$you have$$\sinh(x^2)=x^2+\frac{x^6}{3!}+\frac{x^{10}}{5!}+\cdots$$and therefore its $34$th derivative at $0$ is $\frac{34!}{17!}$. And, clearly, $\cosh^{(34)}=\cosh$ and therefore $\cosh^{(34)}(0)=1$.