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 $\displaystyle y' = e^{x} + xe^{x}$