Let f : R → R be a function satisfying $f(x+y)=f(x)+\lambda x y+3 x^2 y^2$  for all  $x, y \in R$. If f(3) = 4 and f(5) = 52, then f'(x) is equal to

$-10 x$

We have,

$f(x+y)=f(x)+\lambda x y+3 x^2 y^2$ for all $x, y \in R$

Putting x = 3 and y = 2, we get

$f(5)=f(3)+6 \lambda+108$

$\Rightarrow 52=4+6 \lambda+108$

$\Rightarrow \lambda=-10$

∴ $f(x+y)=f(x)-10 x y+3 x^2 y^2$

$\Rightarrow \frac{f(x+y)-f(x)}{y}=-10 x+3 x^2 y$

$\Rightarrow \lim\limits_{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim\limits_{y \rightarrow 0}-10 x+3 x^2 y$

$\Rightarrow f'(x)=-10 x$