Question

$y=\sqrt{x+\frac{1}{x}}$
differentiate w.r.t $x$

Answer 1

$y=\sqrt{x+\frac{1}{x}}letx+\frac{1}{x}=t\frac{dt}{dx}=1-\frac{1}{x^2}y=\sqrt{t}\frac{dy}{dx}=\frac{1}{2\sqrt{t}}\frac{dt}{dx}=\frac{1}{2\sqrt{x+\frac{1}{x}}}\times (1-\frac{1}{x^2})$