How do I find the critical points of this function involving e? I have the function:
$$g(x)={{1 \over \sqrt{2 \pi}} \cdot e^{{-(x-2)^2}/2}}$$
Through very tedious differntion, I got to:
$$g'(x) = {{{-(x+2)} \cdot {e^{{-(x-2)^2}/2}}} \over {2 \pi}}$$
Setting $g'(x) =0$ like I was taught in class, I am unable to find the critical points. The farthest I can get is:
$$xe^{u} = 2e^u$$ where u = that annoying exponent.
Something tells me to take the ln of both sides, but I can't get it in a useful form to do so.
How do I proceed? 
 A: The only $x$ that makes that zero is $x=-2$.  Divide both sides by the $e^{\dots}$ part which you know is never zero.
A: If $g(x) =\frac{1}{\sqrt{2\pi}}e^{-(x-2)^2}$, then
$$g'(x)=\frac{1}{\sqrt{2\pi}}e^{-(x-2)^2} \times (-2(x-2))$$
for which $g'$ is zero only at $x=2$.
At $x=2$, we have $g(2)=\frac{1}{\sqrt{2\pi}}$.  
This value is the maximum of $g$.
A: Apparently WolframAlpha isn't enough to show you that your derivative is incorrect, so let me help you see for yourself.
We're starting out with this (edited for clarity)
$$g(x)=2\pi^\frac{-1}{2} * e^\frac{-(x-2)^2}{2}$$
I'm going to pull out the constant $2\pi^\frac{-1}{2}$ so now we have
$$2\pi^\frac{-1}{2}g(x)=e^\frac{-(x-2)^2}{2}$$
Using the chain rule, we get
$$2\pi^\frac{-1}{2}g'(x)=(\frac{-(x-2)^2}{2})'e^\frac{-(x-2)^2}{2}$$
Simplifying the derivative:
$$(\frac{-(x-2)^2}{2})' = (\frac{-2(x-2)}{2}) = -(x-2) = 2-x$$
Therefore
$$2\pi^\frac{-1}{2}g'(x)=(2-x)e^\frac{-(x-2)^2}{2}$$
And the only time that $g'(x)=0$ is when $x = +2$, so the critical point is
$$g(2) = 2\pi^\frac{-1}{2}$$
A: If you have a complicated function  $g(x)$ involving a lot of products and exponentials ask John Napier for help. He would tell you to make it even more complicated $f(x)=\log(g(x))$. Then by the chain rule 
$f'(x)=\frac{g'(x)}{g(x)}$ or $g'(x)=g(x)\cdot f'(x)$.
What would be the gain? Well $\log(g(x))$ might be easier to differentiate.
In your case 
$\log(g(x))=\log(1/\sqrt{2\pi})-(x-2)^2/2$ and its derivative is $-(x-2)$
so that $g'(x)=g(x)\cdot(2-x)$
