Answer
$$y'=\frac{(2x+1)\sqrt{x(x+1)}}{2x(x+1)}$$
Work Step by Step
$$y=\sqrt{x(x+1)}$$
Logarithmic differentiation helps to simplify the differentiation of complex functions by using the algebraic properties of logarithm. In detail,
1) Take the natural logarithm of both sides and simplify:
$$\ln y=\ln\sqrt{x(x+1)}=\ln\Big(\sqrt x\times\sqrt{x+1}\Big)$$
Recall that $\ln(A\times B)=\ln A+\ln B$: $$\ln y=\ln\sqrt x+\ln\sqrt{x+1}$$
2) Then we take the derivative of both sides with respect to $x$ and remember that $(\ln x)'=1/x$:
$$\frac{1}{y}\times y'=\frac{1}{\sqrt x}(\sqrt x)'+\frac{1}{\sqrt{x+1}}(\sqrt{x+1})'$$ $$\frac{y'}{y}=\frac{1}{\sqrt x}\times\frac{1}{2\sqrt x}+\frac{1}{\sqrt{x+1}}\times\frac{1}{2\sqrt{x+1}}(x+1)'$$ $$\frac{y'}{y}=\frac{1}{2x}+\frac{1}{2(x+1)}=\frac{x+1+x}{2x(x+1)}=\frac{2x+1}{2x(x+1)}$$
3) Solve for $y'$: $$y'=\frac{2x+1}{2x(x+1)}\times y$$
4) Finally, substitute for $y=\sqrt{x(x+1)}$
$$y'=\frac{(2x+1)\sqrt{x(x+1)}}{2x(x+1)}$$
