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

$$y'=\frac{3x+4}{(2\ln2)x(x+1)}$$
$$y=\log_2\Big(\frac{x^2e^2}{2\sqrt{x+1}}\Big)$$ Recall Theorem 7: $$\frac{d}{dx}\log_au=\frac{1}{u\ln a}\frac{du}{dx}$$ So the derivative of $y$ is: $$y'=\frac{1}{\frac{x^2e^2}{2\sqrt{x+1}}\times\ln2}\Big(\frac{x^2e^2}{2\sqrt{x+1}}\Big)'=\frac{2\sqrt{x+1}}{x^2e^2\ln2}\Big(\frac{x^2e^2}{2\sqrt{x+1}}\Big)'$$ We examine separately $(\frac{x^2e^2}{2\sqrt{x+1}})'$: $$\Big(\frac{x^2e^2}{2\sqrt{x+1}}\Big)'=\frac{(x^2e^2)'(2\sqrt{x+1})-(x^2e^2)(2\sqrt{x+1})'}{4(x+1)}$$ $$=\frac{2e^2x(2\sqrt{x+1})-(x^2e^2)\Big(\frac{2}{2\sqrt{x+1}}\Big)}{4(x+1)}$$ $$=\frac{4e^2x\sqrt{x+1}-\frac{x^2e^2}{\sqrt{x+1}}}{4(x+1)}=\frac{\frac{4e^2x(x+1)-x^2e^2}{\sqrt{x+1}}}{4(x+1)}$$ $$=\frac{4x^2e^2+4xe^2-x^2e^2}{4(x+1)\sqrt{x+1}}=\frac{3x^2e^2+4xe^2}{4(x+1)\sqrt{x+1}}$$ $$=\frac{xe^2(3x+4)}{4(x+1)\sqrt{x+1}}$$ Therefore, $$y'=\frac{2\sqrt{x+1}}{x^2e^2\ln2}\times\frac{xe^2(3x+4)}{4(x+1)\sqrt{x+1}}=\frac{1}{x\ln2}\times\frac{3x+4}{2(x+1)}$$ $$y'=\frac{3x+4}{(2\ln2)x(x+1)}$$