Answer
$f^{\prime}(x)=(2x+1)^{-0.5}$
Work Step by Step
Let $u(x)=x^{0.5},\qquad v(x)=2x+1$
$\displaystyle \frac{du}{dx}=0.5x^{-0.5}, \quad \frac{dv}{dx}=2$
Then,$\quad y=f(x)=u(v(x)$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=0.5v^{-0.5}=0.5(2x+1)^{-0.5}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=0.5(2x+1)^{-0.5}\cdot 2=(2x+1)^{-0.5}$
