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The graph of the function y=e^(x^2/2) isn't the same as the graph of the function y=e^(x) is it because that the first expression wants a positive value of all x?
And does anyone know d/dx (x^21)e^(x^2/2)?
 one year ago
 one year ago
The graph of the function y=e^(x^2/2) isn't the same as the graph of the function y=e^(x) is it because that the first expression wants a positive value of all x? And does anyone know d/dx (x^21)e^(x^2/2)?
 one year ago
 one year ago

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student92Best ResponseYou've already chosen the best response.0
Both \[f(x)=e^{(x^2/2)}\]\[y=e^{x}\]may have negative value for x. Following links provide the graph of each function. f(x) > http://www.wolframalpha.com/input/?i=y%3De^%28x^2%2F2%29 g(x) > http://www.wolframalpha.com/input/?i=y%3De^%28x%29 Note something: 1) ^2 cause concave up. 2)  sign cause mirror image of the + graphs. Hence g(x) is a mirror image of e^x , whereas f(x) has a concave, but it is down due to the negative sign.
 one year ago

student92Best ResponseYou've already chosen the best response.0
On the second question First of all, multiply/rearrange the function. \[f(x)=(x^21)(e^{x^2/2})=x^2 e^{x^2/2}e^{x^2/2}\]Next, we use chain rule.\[f'(x)=2xe^{x^2/2}+x^2(2x/2)e^{x^2/2}+()(2x/2)e^{x^2/2}\]\[f'(x)=2xe^{x^2/2}x^3e^{x^2/2}+xe^{x^2/2}\]\[f'(x)=e^{x^2/2}(2xx^3+x)\]\[f'(x)=e^{x^2/2}(3xx^3)\]
 one year ago

student92Best ResponseYou've already chosen the best response.0
P.S. g(x) I mentioned on the first response is y=e^x
 one year ago

JingleBellsBest ResponseYou've already chosen the best response.0
This is what I meant and I graphed and expected to see the same graph...
 one year ago

student92Best ResponseYou've already chosen the best response.0
\[\frac{1}{\sqrt{e^{x^2}}} \neq \frac{1}{e^x}\] I think you mistaken \[\frac{1}{\sqrt{e^{x^2}}} \text{as} \frac{1}{\sqrt{e^{2x}}}\]See? say for example e=2.7 and let x=2. If you subtitle those values to what you meant before, the left hand side (LHS) will be diferrent to right hand side (RHS).
 one year ago

JingleBellsBest ResponseYou've already chosen the best response.0
of course haha it's been a long day thanks! I was just starting to freak out...
 one year ago

student92Best ResponseYou've already chosen the best response.0
LOL. You just need to take a fresh air. Observe that the ^2 here is for x only; not (e^x) altogether. \[e^{(x^2)} \neq (e^x)^2 \]
 one year ago

JingleBellsBest ResponseYou've already chosen the best response.0
Good idea! Going for a run and pump some more blood into my head. Thanks!
 one year ago

opendiscussBest ResponseYou've already chosen the best response.0
student 92.... i do not think e(x^2)≠(e^x)^2 They are both equal....
 one year ago

student92Best ResponseYou've already chosen the best response.0
let just say \( x=1 \). \[e(x^2)=e(1^2)=e \]\[(e^x)^2=(e^1)^2=e^2 \]Are they still the same?
 one year ago
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