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

$$y'=\Big(\frac{1}{x}+\frac{x}{(x^2+1)}-\frac{2}{3(x+1)}\Big)\frac{x\sqrt{x^2+1}}{(x+1)^{2/3}}$$
$$y=\frac{x\sqrt{x^2+1}}{(x+1)^{2/3}}$$ 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: $$\ln y=\ln\Big(\frac{x\sqrt{x^2+1}}{(x+1)^{2/3}}\Big)$$ Recall that $\ln(A\times B)=\ln A+\ln B$ and $\ln(A/B)=\ln A-\ln B$: $$\ln y=\ln(x\sqrt{x^2+1})-\ln(x+1)^{2/3}$$ $$\ln y=\ln x+\ln\sqrt{x^2+1}-\ln(x+1)^{2/3}$$ 2) Then we take the derivative of both sides with respect to $x$ and remember that $(\ln x)'=1/x$: $$\frac{1}{y}\times y'=\frac{1}{x}+\frac{1}{\sqrt{x^2+1}}(\sqrt{x^2+1})'-\frac{1}{(x+1)^{2/3}}((x+1)^{2/3})'$$ $$\frac{y'}{y}=\frac{1}{x}+\frac{1}{\sqrt{x^2+1}}\times\frac{1}{2\sqrt{x^2+1}}(x^2+1)'-\frac{1}{(x+1)^{2/3}}\times\frac{2}{3}(x+1)^{-1/3}$$ $$\frac{y'}{y}=\frac{1}{x}+\frac{(x^2+1)'}{2(x^2+1)}-\frac{2}{3(x+1)^{2/3}(x+1)^{1/3}}$$ $$\frac{y'}{y}=\frac{1}{x}+\frac{2x}{2(x^2+1)}-\frac{2}{3(x+1)}$$ $$\frac{y'}{y}=\frac{1}{x}+\frac{x}{(x^2+1)}-\frac{2}{3(x+1)}$$ 3) Solve for $y'$: $$y'=\Big(\frac{1}{x}+\frac{x}{(x^2+1)}-\frac{2}{3(x+1)}\Big)\times y$$ 4) Finally, substitute for $y=\frac{x\sqrt{x^2+1}}{(x+1)^{2/3}}$ $$y'=\Big(\frac{1}{x}+\frac{x}{(x^2+1)}-\frac{2}{3(x+1)}\Big)\frac{x\sqrt{x^2+1}}{(x+1)^{2/3}}$$