## University Calculus: Early Transcendentals (3rd Edition)

$$y'=\frac{5x\sqrt{(x+1)^{10}}}{(x+1)(2x+1)\sqrt{(2x+1)^5}}$$
$$y=\sqrt{\frac{(x+1)^{10}}{(2x+1)^5}}$$ Logarithmic differentiation helps to simplify the differentiation of otherwise complex functions by using the algebraic properties of logarithm. In detail, 1) Take the natural logarithm of both sides and simplify: $$\ln y=\ln\Big(\sqrt{\frac{(x+1)^{10}}{(2x+1)^5}}\Big)=\ln\Big(\frac{\sqrt{(x+1)^{10}}}{\sqrt{(2x+1)^5}}\Big)$$ Recall that $\ln(A/B)=\ln A-\ln B$: $$\ln y=\ln\sqrt{(x+1)^{10}}-\ln\sqrt{(2x+1)^5}$$ 2) Then we take the derivative of both sides with respect to $x$ and remember that $(\ln x)'=1/x$: $$\frac{1}{y}\times y'=\frac{1}{\sqrt{(x+1)^{10}}}(\sqrt{(x+1)^{10}})'-\frac{1}{\sqrt{(2x+1)^{5}}}(\sqrt{(2x+1)^{5}})'$$ $$\frac{y'}{y}=\frac{1}{\sqrt{(x+1)^{10}}}\times\frac{1}{2\sqrt{(x+1)^{10}}}((x+1)^{10})'-\frac{1}{\sqrt{(2x+1)^{5}}}\times\frac{1}{2\sqrt{(2x+1)^{5}}}((2x+1)^5)'$$ $$\frac{y'}{y}=\frac{((x+1)^{10})'}{2(x+1)^{10}}-\frac{((2x+1)^5)'}{2(2x+1)^5}$$ $$\frac{y'}{y}=\frac{10(x+1)^9}{2(x+1)^{10}}-\frac{5(2x+1)^4(2x+1)'}{2(2x+1)^5}$$ $$\frac{y'}{y}=\frac{5}{x+1}-\frac{5\times2}{2(2x+1)}=\frac{5}{x+1}-\frac{5}{2x+1}$$ $$\frac{y'}{y}=\frac{10x+5-5x-5}{(x+1)(2x+1)}=\frac{5x}{(x+1)(2x+1)}$$ 3) Solve for $y'$: $$y'=\frac{5x}{(x+1)(2x+1)})\times y$$ 4) Finally, substitute for $y=\sqrt{\frac{(x+1)^{10}}{(2x+1)^5}}$ $$y'=\frac{5x}{(x+1)(2x+1)}\times\sqrt{\frac{(x+1)^{10}}{(2x+1)^5}}=\frac{5x\sqrt{(x+1)^{10}}}{(x+1)(2x+1)\sqrt{(2x+1)^5}}$$