Question

Please explain it as much as possible​

Answer 1

`f(x)=\ln\left( x+√(x^2-1) \right)\implies f(x)=\ln\left( x+(x^2-1)^{(1)/(2)} \right) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{df}{dx}\implies \stackrel{ \textit{\large chain rule}\qquad }{\cfrac{1}{ x+(x^2-1)^{(1)/(2)}}\left[1~~ + ~~\stackrel{ \textit{\small chain rule} }{\cfrac{1}{2}(x^2-1)^{-(1)/(2)}(2x)} \right]}`

`\cfrac{1}{ x+(x^2-1)^{(1)/(2)}}\left[1~~ + ~~(x^2-1)^{-(1)/(2)}(x) \right]\implies \cfrac{1}{ x+(x^2-1)^{(1)/(2)}}\left[1~ + ~\cfrac{x}{(x^2-1)^{(1)/(2)}} \right] \\\\\\ \cfrac{1}{ x+(x^2-1)^{(1)/(2)}}\left[\cfrac{(x^2-1)^{(1)/(2)}+x}{(x^2-1)^{(1)/(2)}} \right]\implies \cfrac{1}{ x+(x^2-1)^{(1)/(2)}}\left[\cfrac{x+(x^2-1)^{(1)/(2)}}{(x^2-1)^{(1)/(2)}} \right] \\\\\\ \cfrac{1}{(x^2-1)^{(1)/(2)}}\implies \cfrac{1}{√(x^2-1)}`