Question
Prove that function $f ( x )=x-\frac{1}{x}, x \in R$ and $x \neq 0$ is increasing function
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$\left(\frac{d^3 y}{d x^3}\right)^2=\sqrt[5]{1+\frac{d y}{d x}}$
$y=\sin x, x=0, x=\frac{\pi}{2}$
$\sqrt{x^2+2 x+5}$