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.

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.
