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]^{(1)/(2)}\implies \cfrac{(x+1)^{(5)/(2)}}{(x+2)^(10)} \\\\\\ \textit{using the quotient rule} \\\\\\ \cfrac{(5)/(2)(x+1)^{(3)/(2)}\cdot 1\cdot (x+2)^(10)-(x+1)(5)/(2)\cdot 10(x+2)^9\cdot 1}{\left[ (x+2)^(10) \right]^2}`


`\bf \cfrac{(5(x+1)√(x+1)(x+2)^(10)-2(x+1)^2√(x+1)\cdot 10(x+2)^9)/(2)}{(x+2)^(20)} \\\\\\ \cfrac{5(x+1)√(x+1)(x+2)^(10)-20(x+1)^2√(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)√(x+1)(x+2)-20(x+1)^2√(x+1)}{2(x+2)^(11)} \\\\\\ \textit{common factor atop of }5(x+1)√(x+1) \\\\\\ \cfrac{5(x+1)√(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{(5(x+1)√(x+1)\left[ (x+2)-4(x+1) \right])/(2(x+2)^(11))}{\sqrt{((x+1)^5)/((x+2)^(20))}} \\\\\\ \cfrac{(5(x+1)√(x+1)\left[ (x+2)-4(x+1) \right])/(2(x+2)^(11))}{\frac{√((x+1)^5)}{\sqrt{(x+2)^(20)}}} \\\\\\ \cfrac{5(x+1)√(x+1)\left[ (x+2)-4(x+1) \right]}{2(x+2)^(11)}\cdot \cfrac{\sqrt{(x+2)^(20)}}{√((x+1)^5)}`


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