thank you i solved the problem
I icyhot2590 New member Joined Mar 18, 2007 Messages 22 Mar 18, 2007 #1 thank you i solved the problem
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 18, 2007 #2 Re: Find y' if y =log5 (sqrtx)/(x+1) Hello, icyhot2590! Is that square root only on the x\displaystyle xx ? Find y′\displaystyle y'y′ if \(\displaystyle y \:=\:\log_5\L\left(\frac{\sqrt{x}}{x\,+\,1}\right)\) Click to expand... We have: y = 12⋅log5(x) − log5(x + 1)\displaystyle \:y \;=\;\frac{1}{2}\cdot\log_5(x) \,- \,\log_5(x\,+\,1)y=21⋅log5(x)−log5(x+1) Then: \(\displaystyle \L\:y'\;=\;\frac{1}{2}\cdot\frac{1}{\ln 5}\cdot\frac{1}{x}\,-\,\frac{1}{\ln5}\cdot\frac{1}{1\,+\,x}\) . . which equals the answer you got.
Re: Find y' if y =log5 (sqrtx)/(x+1) Hello, icyhot2590! Is that square root only on the x\displaystyle xx ? Find y′\displaystyle y'y′ if \(\displaystyle y \:=\:\log_5\L\left(\frac{\sqrt{x}}{x\,+\,1}\right)\) Click to expand... We have: y = 12⋅log5(x) − log5(x + 1)\displaystyle \:y \;=\;\frac{1}{2}\cdot\log_5(x) \,- \,\log_5(x\,+\,1)y=21⋅log5(x)−log5(x+1) Then: \(\displaystyle \L\:y'\;=\;\frac{1}{2}\cdot\frac{1}{\ln 5}\cdot\frac{1}{x}\,-\,\frac{1}{\ln5}\cdot\frac{1}{1\,+\,x}\) . . which equals the answer you got.
