Implicit derivative of e^(xy^2) = x - y

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I have to agree with you - I really prefer the d/dx and dy/dx notation just for keeping things easier to follow. (And with my writing, I'm sure the y' would quickly be mistaken with a y^1 :) )

mlsboyscout

I've no problems with the actual process outlined but what I don't quite understand is how do you go about graphing a function like this to begin with? Example: when x = 1, substituted into the original function you get e^y^2 = 1 - y, but how would you go about solving this equation for Y to find the coordinate pair for that point??? Taking natural logs doesn't seem to work as you're simply left with ln(1-Y) on the right hand side, with the Y here locked in to the logarithm.

damianreid

hey khan i have a question if both y' are on the same side like y' + y' can you add that to get 2y' ??? is that possible ?

ihatedrake

So Thanks, now i can take a deep breath.

cluelessarn

It's not y', you are not multiplying, you're adding. So it will be
( expression 'times' y' + y' )and then you have to factorize to get ( expression + 1 ) y'

behnamasid

at around 4:15 you decided to get the yprimes on one side. However, ypirme plus yprime is 2yprime right?

mattreisenweber

imagine taking the anti derivative of that final equation...

caleborp

when you where adding y' where gone 2nd one ? (4:10)

sum should be 2y' i think

kjezier

what if we had the same function but with out the square, will D[xy] just disappear ??

S_P_S-tm

Implicit derivative of e^(xy^2) = x - y

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