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JingleBells
 3 years 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)?
JingleBells
 3 years 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)?

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student92
 3 years ago
Best ResponseYou've already chosen the best response.0Both \[f(x)=e^{(x^2/2)}\]\[y=e^{x}\]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.

student92
 3 years ago
Best ResponseYou've already chosen the best response.0On 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.\[f'(x)=2xe^{x^2/2}+x^2(2x/2)e^{x^2/2}+()(2x/2)e^{x^2/2}\]\[f'(x)=2xe^{x^2/2}x^3e^{x^2/2}+xe^{x^2/2}\]\[f'(x)=e^{x^2/2}(2xx^3+x)\]\[f'(x)=e^{x^2/2}(3xx^3)\]

student92
 3 years ago
Best ResponseYou've already chosen the best response.0P.S. g(x) I mentioned on the first response is y=e^x

JingleBells
 3 years ago
Best ResponseYou've already chosen the best response.0This is what I meant and I graphed and expected to see the same graph...

student92
 3 years ago
Best 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).

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

student92
 3 years ago
Best ResponseYou've already chosen the best response.0LOL. You just need to take a fresh air. Observe that the ^2 here is for x only; not (e^x) altogether. \[e^{(x^2)} \neq (e^x)^2 \]

JingleBells
 3 years ago
Best ResponseYou've already chosen the best response.0Good idea! Going for a run and pump some more blood into my head. Thanks!

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

student92
 3 years ago
Best ResponseYou've already chosen the best response.0let just say \( x=1 \). \[e(x^2)=e(1^2)=e \]\[(e^x)^2=(e^1)^2=e^2 \]Are they still the same?
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