taking the derivative of e^(x^2)

• Nov 13th 2012, 09:35 PM
kingsolomonsgrave
taking the derivative of e^(x^2)
I'm not sure how to take the derivative of \$\displaystyle e^{x^2}\$

according to this: Attachment 25704

it is \$\displaystyle e^{x^2}*2x\$ (using l'hospital's rule) but how did they get that? when I plug it into maple I get \$\displaystyle 2e^{x^2}x ln(e) \$ which is the same as what is above (given that ln(e) =1) but I don't know why that should be the answer.
• Nov 13th 2012, 10:06 PM
earboth
Re: taking the derivative of e^(x^2)
Quote:

Originally Posted by kingsolomonsgrave
I'm not sure how to take the derivative of \$\displaystyle e^{x^2}\$

according to this: Attachment 25704

it is \$\displaystyle e^{x^2}*2x\$ (using l'hospital's rule <-- ??) but how did they get that? when I plug it into maple I get \$\displaystyle 2e^{x^2}x ln(e) \$ which is the same as what is above (given that ln(e) =1) but I don't know why that should be the answer.

Have a look here: Chain rule - Wikipedia, the free encyclopedia
• Nov 13th 2012, 10:14 PM
kingsolomonsgrave
Re: taking the derivative of e^(x^2)
OH, I did not realize that the chain rule would apply to an exponent that is itself a function of x. thanks

also I meant they were applying l'hospital's rule to the entire expression, not to \$\displaystyle e^{x^2}\$

100th post!(Clapping)