Question

Find the derivative of $y=\sqrt{x^2+1}$.

Answer 1

Here $y=f\left(g\right(x\left)\right)$, where $f\left(u\right)=g\left(x\right)$ and $g\left(x\right)=x^2+1$. Since the derivatives of $f$ and $g$ are
$f^{\prime }\left(u\right)=\frac{1}{2\sqrt{u}}$ and $g^{\prime }\left(x\right)=2x$,
the chain rule gives
$\frac{dy}{dx}=\frac{d}{dx}f\left(g\right(x\left)\right)$
$=f^{\prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)$
$=\frac{1}{2\sqrt{g\left(x\right)}}\cdot g^{\prime }\left(x\right)=\frac{1}{2\sqrt{x^2+1}}\cdot \left(2x\right)=\frac{x}{\sqrt{x^2+1}}$.