Hey help explain this to me Find the derirative of f'(x) f(x+y)=f(x)+f(y)+x^2 y+y^2 x?

Question

f(x) is a function and not a constant. The answer is x^2+1, but I do not know how. The derirative is with respect to x

Answer 1

Hi

Here's my solution:

The functional equation is f(x+y) = f(x)+f(y)+xy(x+y)
If you plug in y = 0, you get f(x) = f(x) + f(0), therefore f(0) = 0. (this will be useful later on)

So how to find f'(x)? This is what I did:
f(x+y) = f(x)+f(y)+xy(x+y)
f(x+y)-f(x) = f(y)+xy(x+y)
[f(x+y)-f(x)]/y = f(y)/y + x(x+y)
Now put limits of y approaching 0 to both sides.
lim y-> 0 [f(x+y)-f(x)]/y = lim y->0 f(y)/y + x(x+y)
Note that on the left hand side, you get the definition of derivative for f(x), namely f'(x).

f'(x) = lim y->0 f(y)/y + x(x+y) = lim y->0 f(y)/y + x^2
In order to evaluate the limit of f(y)/y, you have to use L'Hopital's rule, and since it is shown above that f(y) = 0 and therefore the limit takes the form of 0/0, it is appropiate to use the L'Hopital's rule in here. (for more info: http://en.wikipedia.org/wiki/L%27hopital%27s_rule )

f'(x) = lim y->0 f'(y)/1 + x^2
f'(x) = f'(0) + x^2
We don't know what f'(0) is. But in fact f'(0) can be any number, so the solution is x^2+C, where C is any real number and the function itself is x^3/3 + Cx.