Chain rule with second derivative

[beetle2]

Homework Statement

I trying to find the second derivative of $xe^x$

Homework Equations

chain rule

The Attempt at a Solution

Two find the first derivative I use the chain rule.

$f'(y)g(y)+f(y)g'(y)$

so I get

$e^x+xe^x$

is the second derivative

$e^x+f'(y)g(y)+f(y)g'(y)$

= $e^x+e^x+xe^x$
= $2e^x+xe^x$

regards

 

[rock.freak667]

That should be correct, also, you used the product rule, not the chain rule.

 

[HallsofIvy]

It can be shown that, for any positive integer n, with $f^{(n)}(x)$ indicating the nth derivative, that
$$(fg)^{(n)}(x)= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix}f^{(i)}(x)g^{(n- i)}(x)$$

And, as rock.freak667 said, that is the product rule.

 

[beetle2]

thanks guys