I must say the veracity of the affirmation : The second degree Taylor's polynomial of around is .
My attempt : I didn't use the substitution. Rather I did it the hard way : .
Not sure I've done it right though. Could you second my answer?
( I just plotted the functions and seems to fit pretty well... meaning I made a mistake but I don't see where!)
I see. I get , so evaluated in and divided by and multiplied by gives .
However I get that , evaluated in it gives . Hence . ( I checked what you told me, that the mixed partial derivatives are 0, that is, )
I feel like I'm missing something obvious. I'll go to sleep and come back tomorrow. Thanks for your help CaptainBlack.