Finding the critical points of $\sin(x)/x$ and $\cosh(x^2)$ Could someone help me solve this:

What are all critical points of $f(x)=\sin(x)/x$ and $f(x)=\cosh(x^2)$?

Mathematica solutions are also accepted.
 A: $0$ is the only critical point of $x \mapsto \cosh x^2$ since its derivative $x\mapsto 2x\sinh x^2$ vanishes only at $x=0$. For $f(x)=\frac{\sin x}{x}$ first notice that $0$ is a removable singularity since $\lim_{x \to 0}f(x)=1$, so we can set $f(0)=1$. Then 
$$
f'(x)=\frac{x\cos x-\sin x}{x^2}=\frac{\cos x}{x^2}(x-\tan x) \quad \forall x \ne 0
$$
with 
$$
f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\frac{1-\frac{x^2}{6}-1}{x}=\lim_{x \to 0}\frac{-x}{6}=0.
$$
Thanks to the Intermediate Value Theorem one shows that for every positive integer $n$ the equation $x-\tan x=0$ possesses a unique solution 
$$
x_n \in ((n-\frac{1}{2})\pi,(n+\frac{1}{2})\pi).
$$ 
By symmetry the set of critical points of $f$ is $\{0,\pm x_n: \ n \in \mathbb{N}\}$. 
A: Taken from here :
$x=c$ is a critical point of the function $f(x)$ if $f(c)$ exists and if one of the following are true:


*

*$f'(c) = 0$

*$f'(c)$ does not exist


The general strategy for finding critical points is to compute the first derivative of $f(x)$ with respect to $x$ and set that equal to zero.
$$f(x) = \frac{\sin x}{x}$$ 
Using the quotient rule, we have:
$$f'(x) = \frac{x\cdot \cos x - \sin x \cdot 1}{x^2}$$
$$f'(x) = \frac{x \cos x}{x^2} - \frac{\sin x}{x^2}$$
Dividing through by $x$ for the left terms, we now have:
$$f'(x) = \frac{\cos x}{x} - \frac{\sin x}{x^2}$$
Now set that equal to zero and solve for your critical points. Do the same for $f(x) = \cosh(x^2)$. Don't forget the chain rule!
For $f(x) = \cosh (x^2)$, recall that $\frac{d}{dx} \cosh (x) = \sinh (x)$. So,
$$f'(x) = \sinh(x^2) \cdot \frac{d}{dx} (x^2)$$
$$f'(x) = 2x \sinh(x^2)$$
$$0 = 2x \sinh(x^2) $$
$x = 0$ is your only critical point along the reals.
A: As mentioned, there is a solution in each interval $(k \pi, (k+1)\pi)$.  This solution can't be expressed in "closed form", but there is a series in negative powers of $k$:
$$x = (k+1/2)\pi -{\frac {1}{k\pi }}+{\frac {1}{2 \pi \,{k}^{2}}}-{\frac {3\,{
\pi }^{2}+8}{12{\pi }^{3}{k}^{3}}}+{\frac {{\pi }^{2}+8}{8{\pi }^{3}
{k}^{4}}}-{\frac {15\,{\pi }^{4}+240\,{\pi }^{2}+208
}{{240 \pi }^{5}{k}^{5}}}+{\frac {3\,{\pi }^{4}+80\,{\pi 
}^{2}+208}{96{\pi }^{5}{k}^{6}}}+\ldots $$
It looks to me like this  converges for $k \ge 1$ (I'm not sure about $k=1$).
