$f(x) + f(y) + f(x)f(y) = 1$For all values of $x,y \in R$$f'(x) = ?$ Tuesday, December 21, 2021 Putting y = x we have,$2f(x) + {\left[ {f(x)} \right]^2} = 1$$2f(x) + {\left[ {f(x)} \right]^2} + 1 = 1 + 1$${[f(x) + 1]^2} = 2$$f(x) = - 1 \pm \sqrt 2 $Since f(x) is a constant function, $f'(x) = 0$.

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