Value of function

hmmmmmmmm

puttin y=0 u get f(x) =0

then frm d given limit u get
f'(0)=0

nw diffrntiate d given equation

ie

f(x+y)=f(x)+f(y)+x^2y

n then after diffrntiatiing put y=0

u get
f'(x) =x^2

so f'(10)=100

:D

cheero~

We have
f^\prime(10)=\lim_{\Delta x\to 0}\dfrac{f(10+\Delta x)-f(10)}{\Delta x}
Using the given relation, we get
f(10+\Delta x)=f(10)+f(\Delta x)+100\Delta x
Therefore,
\dfrac{f(10+\Delta x)-f(10)}{\Delta x}=\dfrac{f(\Delta x)}{\Delta x}+100
Taking limit, and using the fact that the limit of f(x)/x =0 as x->0, we get f'(10)=100