Question

If $y=\sqrt{x+1}+\sqrt{(x^2+1{)}^2}$ Then $\frac{dy}{dx}$?

Answer 1

$y=\sqrt{x+1}+\sqrt{(x^2+1{)}^2}$

$\therefore y=\sqrt{x+1}+x^2+1$

$\therefore \frac{dy}{dx}=\frac{\left(1\right)}{2\sqrt{x+1}}+2x$

$\therefore \frac{dy}{dx}=2x+\frac{1}{2\sqrt{x+1}}$

$\therefore \frac{dy}{dx}=2x+\frac{1}{2\sqrt{x+1}}$.