Question

Help please! What is the derivative of
y = e^(1+ln(x)) with steps please!!!

Answer 1

Chain rule! this not some fake question! you know how to do it!

Answer 2

i'm suspicious about it lol

Answer 3

agree with @xapproachesinfinity

Answer 4

Actually I am not getting it correct and I can't figure out why!

Answer 5

seriously, please help. I can't get it right!

Answer 6

Question: is it not that e^ (a+b) = (e^a )(e^b)?

Answer 7

okay anyways!
\(\Large \rm (f(g(x))'=f'(g(x))*g'(x)\)
chain rule says take the derivative of the outer with respect to the inner variable g(x) multiplied by the derivative of the inner function with respect to x
so \(\huge y=e^{1+lnx}=e*e^{lnx}\)
so \(\large y'=e(e^{lnx})'=e*e^{lnx}*(lnx)'=e*e^{lnx}*1/x\)

Answer 8

writing it in another way
\(\huge \rm y'=\frac{e^{1+lnx}}{x}\)

Answer 9

One more question: is it not that \(e^{ln(x)}=x\)

Answer 10

hhhh i'm not paying attention lol

Answer 11

so the final product is e

Answer 12

:)

Answer 13

i was focused on derivative i didn't realize it was e^lnx

Answer 14

ok you guys, I get it now! I didn't remember about the derivative of e^lnx either! That's the whole problem! I don't know who to medal for this. If I medal one of you, the one I medall needs to medal the other guy!!! Thank you both so much!!!!!

Answer 15

writing things neatly
\(\huge y'=\frac{e*x}{x}=e\)

Answer 16

eh don't worry about medals I'm fine!
the derivative of e^x is just e^x

Answer 17

Ty for medaling, @Loser66 . what a good sport you are! It's beeen awhile with the e^ something; I forgot! ugh! Thank you both a;gain, so much!!!!

Answer 18

eh this is stupid of me from the start e^lnx=x
so the function is f(x)=ex
f'(x)=e
i'm dumb lol

Answer 19

Thanks for compliment. I can see other's problem only, not mine. I am still struggling with my problem. If you guys can, please, help

Answer 20

welcome^_^

Answer 21

your problem are tough, I forgot how to do those stuff.