A maths problem

isnt that xye^xy. I think not too sure as i did the differentiation of the expontial like in november and i forgot it straight after the A2 level maths exam in january lol. errr actually i just thought about it and i think you use chain rule i think lol.so differentiate the xy then times that by the differentiation of e^xy which is xye^xy, im not too sure though sorry
 
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No it's not. You've got to differentiate with respect to x by the way.

It's not differentiating e, differentiating with respect to x.
 
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The answer is ye^(xy). You have to use the chain rule:

First, differentiate the "argument" of the function (in this case, xy) to give y. Then, differentiate the function itself, leaving the argument the same (and as you should know, the derivative of an exponential is still an exponential). Multiply the two answers to give ye^(xy).
 
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I got an E at AS level, but am getting A's and B's at A2, mad world eh?

Thanks Napalmbrain, I'll look into it.
 
lolz said:
math sucks. you'll never need to know half the **** they teach you.

I'm sure you'll be very surprised when you are an adult.
 
damn, what level of calc are you in? Or is it just a diff. equations class?

I was about to say it's the same as e^x * e^y, but then I was like, wait, that's e^(x+y), haha. I'm in calc 3 right now. =)

All I can think of, which I'm sure you've already thought of, is to use (e^x)^y and go from there. Go where? I'm not sure.
 
FR. said:
I'm sure you'll be very surprised when you are an adult.
no he wont, even my math teacher says she barely uses half the stuff we learn, and its the easier more basic stuff, we learn all this **** because its better for getting into colleges, which leads into your life, if we all had to only go up to Algebra, then how would colleges decide who they want and who they dont? the more you can do proves how smart you are.

get it?
 
wow. i didnt even read the whole thread and got a headache! i HATE math so much lol. however i must leave now to go to night school for algebra 2 :(
 
All I can think of, which I'm sure you've already thought of, is to use (e^x)^y and go from there. Go where? I'm not sure.

Could use the chain rule with that?

start with:
y(e^x)^(y-1)(xe^(x-1))

Dunno how to go from there...

Hate natural logs :( :(
 

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