Given $y(x) = \sqrt{x\sqrt{(x+3)\sqrt{x\sqrt{(x+3)\sqrt{x...}}}}}$
Which value of $\dfrac{dy(1)}{dx} = y'(1)$?
My attempt:
$y^2 = x\cdot \sqrt{(x+3)\cdot y}\\y^4 = x^2\cdot (x+3)\cdot y\\ y^3 = x^3+3x^2$
So trying $y'$:
${3}\cdot y^2\cdot y' = 3x^2+6x\\y'(x) = \dfrac{x^2+2x}{(x^3+3x^2)^{2/3}}\\ y'(1) = \dfrac{3}{2^{4/3}} = \dfrac{3}{\sqrt[3]{16}}$
The answer in my book tells $\dfrac{3}{8}$.
What I did wrong?
