Question

find the derivative of ln square root (x+1)^5/(x+2)^20

Answer 1

$\bf \cfrac{d}{dx}\left[ ln \left[\sqrt{\cfrac{(x+1)^5}{(x+2)^{20}}} \right]\right]\\\ -----------------------------\\\\ \sqrt{\cfrac{(x+1)^5}{(x+2)^{20}}}\implies \left[ \cfrac{(x+1)^5}{(x+2)^{20}} \right]^{\frac{1}{2}}\implies \cfrac{(x+1)^{\frac{5}{2}}}{(x+2)^{10}} \\\\\\ \textit{using the quotient rule} \\\\\\ \cfrac{\frac{5}{2}(x+1)^{\frac{3}{2}}\cdot 1\cdot (x+2)^{10}-(x+1)\frac{5}{2}\cdot 10(x+2)^9\cdot 1}{\left[ (x+2)^{10} \right]^2}$

$\bf \cfrac{\frac{5(x+1)\sqrt{x+1}(x+2)^{10}-2(x+1)^2\sqrt{x+1}\cdot 10(x+2)^9}{2}}{(x+2)^{20}} \\\\\\ \cfrac{5(x+1)\sqrt{x+1}(x+2)^{10}-20(x+1)^2\sqrt{x+1}(x+2)^9}{2(x+2)^{20}} \\\\\\ \textit{common factor of }(x+2)^{9}\textit{ on top and bottom}\\ \\\\\\ \cfrac{5(x+1)\sqrt{x+1}(x+2)-20(x+1)^2\sqrt{x+1}}{2(x+2)^{11}} \\\\\\ \textit{common factor atop of }5(x+1)\sqrt{x+1} \\\\\\ \cfrac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]}{2(x+2)^{11}}$

now, that's just the derivative of the funtion inside ln()
bear in mind that $\bf \cfrac{d}{dx}ln[f(x)]\implies \cfrac{f'(x)}{f(x)}$

thus, let us give it the denominator to that

$\bf \cfrac{\frac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]}{2(x+2)^{11}}}{\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}} \\\\\\ \cfrac{\frac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]}{2(x+2)^{11}}}{\frac{\sqrt{(x+1)^5}}{\sqrt{(x+2)^{20}}}} \\\\\\ \cfrac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]}{2(x+2)^{11}}\cdot \cfrac{\sqrt{(x+2)^{20}}}{\sqrt{(x+1)^5}}$

$\bf \cfrac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]\cdot (x+2)^{10}}{2(x+2)^{11}\cdot (x+1)^2\sqrt{x+1}} \\\\\\ \textit{now we take the common factors atop and bottom of}\\\\ (x+1)\sqrt{x+1}(x+2)^{10} \\\\\\ thus\implies \cfrac{5\left[ (x+2)-4(x+1) \right]}{2(x+2)(x+1)}$

yeah, we can simplify the numerator only, since the denominator is a cubic

so... $\bf \cfrac{5\left[ (x+2)-4(x+1) \right]}{2(x+2)(x+1)}\implies \cfrac{-5(3x+2)}{2(x+2)(x+1)}$