Question

Differentiate using the chain rule

Answer 1

\[y=xe^{-x^{2}}\]

Answer 2

I realize you need to do the product rule first, then use the chain rule for part of that, can someone please work through the whole problem and explain their steps? It is more confusing for me when you ask me questions about it to try to explain, i understand it the best when you just show the steps

Answer 3

@mathmale @SithsAndGiggles @xapproachesinfinity

Answer 4

The derivative of y is y'(x):
\[\Large y'(x)=\frac{ \delta }{ \delta x }\left( \frac{ x }{ e^{x^{2}} } \right)\]
Use the product rule,
\[\frac{ d }{ dx }(u v) = v \frac{ du }{ dx }+u \frac{ dv }{ dx }, where~u=e^{-x^{2}}~and~v=x\]
\[\Large y'(x)=x \frac{ d }{ dx }\left( e^{-x^2} \right)+\frac{ \frac{ d }{ dx }(x) }{ e^{x^2} }\]
Using the chain rule,
\[\Large \frac{ d }{ dx }(e^{-x^2})=\frac{ de^u }{ du }\frac{ du }{ dx },~where~u=-x^2~and~\frac{ d }{ du }(e^u)=e^u \]
\[\Large y'(x)=\frac{ \frac{ d }{ dx } (x)}{ e^{x^2} }+\frac{ \frac{ d }{ dx }(-x^2) }{ e^{x^2} }x\]

Answer 5

Ok... I think that's what i'm doing, but i'm not sure I completely followed on that last step
Can you tell me what I'm doing wrong here?

Answer 6

i'm not sure where you get the second term at the second line from.

Answer 7

Second line of work or second line of writing?
On the second line of writing, i was breaking it down using the product rule
Third line, I simplified x prime to 1, leaving the e^-x^2, .... then things got ugly bc i'm not quite sure what I'm doing with the chain rule there
So I guess my real question is... what's the derivative of \[e^{-x^2}\]

Answer 8

that also requires the chain rule, and it will result in \(\Large -\frac{ 2x }{ e^{x^2} }\)