We have, $$y = \left(\frac{\sqrt{x+1}}{\sqrt{x}}\right)^3$$ $$ \implies y = \left(1 + \frac{1}{x}\right)^{3/2}$$ Differentiating both sides w.r.t. $x$ (Using chain rule) $$\implies \frac{dy}{dx} = \frac32\left(1 + \frac{1}{x}\right)^{3/2 -1 }\cdot \dfrac{d}{dx}\left(1 + \frac{1}{x}\right)$$ $$ \implies \frac{dy}{dx} = \frac32\left(1 + \frac{1}{x}\right)^{1/2}\cdot \frac{-1}{x^2}$$ $$ \implies \boxed{\frac{dy}{dx} = \frac{-3}{2x^2}\sqrt{1 + \frac{1}{x}}}$$ What is the mistake in my work?
Graphs of $f'(x)$ and $\frac{-3}{2x^2}\sqrt{1 + \frac{1}{x}}$ are not same.