請教我這題微積分!!!謝謝

I remember this question is a pastpaper question.
f(x+y)=f(x)+f(y)+x^2y+xy^2
when x=y=0
f(0+0)=f(0)+f(0)+[(0)^2](0)+(0)(0)^2
f(0)=f(0)+f(0)
f(0)=f(0) - f(0)
f(0)=0

f(x+y)=f(x)+f(y)+x^2y+xy^2
f(x+y)-f(x)=f(y)+x^2y+xy^2
[f(x+y)-f(x)]/y=[f(y)+x^2y+xy^2]/y
lim(y->0) [f(x+y)-f(x)]/y=lim(y->0) [f(y)+x^2y+xy^2]/y
f ' (x)=lim(y->0) [f(y)+x^2y+xy^2]/y (defintion: f ' (x)=lim(y->0) [f(x+y)-f(x)]/y)
f ' (x)=lim(y->0) [f(y)+(x^2)(0)+x(0)^2]/y
f ' (x)=lim(y->0) [f(y)/y]
f ' (x)=lim(x->0) [f(x)/x] (variable y can be changed to variable x)
f ' (x)=1

f ' (0)=1

2013-07-04 10:13:48 補充：
the above answer is mistaken
f ' (x)=lim(y->0) [f(y)+x^2y+xy^2]/y
f ' (x)=lim(y->0) [f(y)/y+(x^2)+xy]
f ' (x)=lim(y->0) [f(y)/y+(x^2)+x(0)]
f ' (x)=lim(y->0) [f(y)/y+(x^2)]
f ' (x)=lim(y->0) [f(y)/y] + (x^2)]
f ' (x)=lim(x->0) [f(x)/x] + (x^2)]
f ' (x)=1 + x^2
f ' (0)=1 + (0)^2=1

f(x+y)=f(x)+f(y)+x^2y+xy^2
(f(x+y)-f(x))/y = f(y)/y+x^2+xy → 1 + x^2 when y→0
So, f'(x) = 1+x^2,
f'(0) = 1.


Check:
f'(x) = 1+x^2 and f(0)=0, so f(x) = x+x^3/3.
f(x+y)-f(x)-f(y) = [(x+y)+(x+y)^3/3] - (x+x^3/3) - (y+y^3/3)
= [(x+y)^3-x^3-y^3]/3 = x^2y + xy^2
OK!