2nd Derivative of xe to the x

[Jason76]

\(\displaystyle g(x) = xe^{x} - x\) -x has the derivative of -1.

leads to

\(\displaystyle g'(x) = e^{x} + xe^{x} - 1\)

and

\(\displaystyle g'(x) = e^{x} + xe^{x} - 1\) -1 has the derivative of 0. \(\displaystyle e^{x}\) has the derivative of \(\displaystyle e^{x}\)

leads to

\(\displaystyle g''(x) = e^{x} + e^{x} + xe^{x}\) Adding the two \(\displaystyle e^{x}\) terms leads to \(\displaystyle 2e^{x}\)

leads to

\(\displaystyle g'' (x) = 2e^{x} + xe^{x}\)

So what about the derivative of \(\displaystyle xe^{x}\) (in both the 1st and 2nd derivatives)? Why does it equal \(\displaystyle e^{x} + xe^{x}\)?

 

[Crimson Sunbird]

So what about the derivative of \(\displaystyle xe^{x}\) (in both the 1st and 2nd derivatives)? Why does it equal \(\displaystyle e^{x} + xe^{x}\)?

Product rule: \(\displaystyle \dfrac{\mathrm d}{\mathrm dx}(uv)=u\dfrac{\mathrm dv}{\mathrm dx}+v\dfrac{\mathrm du}{\mathrm dx}\). Here \(\displaystyle u=x\) and \(\displaystyle v=e^x\).

 

[Jason76]

Product rule: \(\displaystyle \dfrac{\mathrm d}{\mathrm dx}(uv)=u\dfrac{\mathrm dv}{\mathrm dx}+v\dfrac{\mathrm du}{\mathrm dx}\). Here \(\displaystyle u=x\) and \(\displaystyle v=e^x\).

Also note that expressions like \(\displaystyle 4e^{x}\) has a derivative of \(\displaystyle 4e^{x}\) and \(\displaystyle 5e^{x}\) has a derivative of \(\displaystyle 5e^{x}\) etc... (they equal themselves). However, \(\displaystyle xe^{x}\) is different.

So we can generalize if \(\displaystyle y = ae^{x}\) then \(\displaystyle y' = ae^{x}\) where a = any real number except 0 (or should I say integer?). We can also say if \(\displaystyle y = e^{x}\) then \(\displaystyle y' = e^{x}\) Finally, because of product rule, if \(\displaystyle y = xe^{x}\) then \(\displaystyle y' = xe^{x} + e^{x}\) which can also be written as \(\displaystyle y' = e^{x} + xe^{x}\) It's hard to see the product rule in action, because they wrote it backwards (though equally correct).

Looking at the product rule in action:

\(\displaystyle y = xe^{x}\)

\(\displaystyle y' = x(e^{x}) + e^{x}(1)\) \(\displaystyle e^{x}\) is the derivative of \(\displaystyle e^{x}\) and 1 is the derivative of x

leads to

\(\displaystyle y' = xe^{x} + e^{x}\) or \(\displaystyle y' = e^{x} + xe^{x}\)