Question

what is 1st and 2nd der.. e^-x^2/2 ?

Answer 1

\[f(x) = e ^{(-x ^{2}/2)}\]\[f'(x) = -xe ^{(-x ^{2}/2)}\]\[f''(x) = -e ^{(-x ^{2}/2)} + x ^{2}e ^{(-x ^{2}/2)} = e ^{(-x ^{2}/2)} (x ^{2} -1)\]

Answer 2

how is it x^2 in the second derv.. ? I know you applied the product rule.. I got x because that what I copied directly from the first derv.. according the product rule? ..
Can you please explain?

Answer 3

and the first derivative too..
I still don't get it..
Can you please explain..?????

Answer 4

To find the first derivative, use the chain rule:
\[d/dx(e^{f(x)}) = e^{f(x)}*f'(x)\]For your example, f(x) = (-x^2/2) and f'(x) = -x
So substituting these into the first equation we get:
\[e^{(-x^{2}/2)}*(-x)\]
To find the second derivative, use the product and chain rule:
"Derivative of the first, times the second, plus the first times the derivative of the second."
\[(-xe^{(-x^{2}/2)})(-x)+(e^{(-x^{2}/2)})(-1)\]
Then simplifying...
\[(x^2e^{(-x^{2}/2)})+(-e^{(-x^{2}/2)})\]
And some more simplifying...
\[e^{(-x^{2}/2)}(x^2-1)\]
Not sure if I explained it in the best way. Let me know if you need more help. Sorry for the delay in answering.