Toán tìm đạo hám của hàm số y= √ x +1 (1/ √ x -1) 09/07/2021 By Athena tìm đạo hám của hàm số y= √ x +1 (1/ √ x -1)
$y=\sqrt{x+1}.\dfrac{1}{\sqrt{x-1}}$ $→y’=(\dfrac{\sqrt{x+1}}{\sqrt{x-1}})’$ $→y’=\dfrac{(\sqrt{x+1})’\sqrt{x-1}-\sqrt{x+1}(\sqrt{x-1})’}{x-1}$ $→y’=\dfrac{\dfrac{\sqrt{x-1}}{2\sqrt{x+1}}-\dfrac{\sqrt{x+1}}{2\sqrt{x-1}}}{x-1}$ $→y’=\dfrac{\dfrac{-1}{\sqrt{x²-1}}}{x-1}$ $→y’=\dfrac{-1}{\sqrt{x²-1}(x-1)}$ Trả lời
$y=\sqrt{x+1}.\dfrac{1}{\sqrt{x-1}}$ $=\dfrac{\sqrt{x+1}}{\sqrt{x-1}}$ $=\dfrac{\sqrt{x^2-1}}{x-1}$ $y’=\dfrac{(\sqrt{x^2-1})'(x-1)-\sqrt{x^2-1}(x-1)’}{(x-1)^2}$ $=\dfrac{\dfrac{(x^2-1)’}{2\sqrt{x^2-1}}.(x-1)-\sqrt{x^2-1}}{(x-1)^2}$ $=\dfrac{\dfrac{x}{\sqrt{x^2-1}}(x-1)-\sqrt{x^2-1}}{(x-1)^2}$ $=\dfrac{\dfrac{x\sqrt{x-1}}{\sqrt{x+1}}-\sqrt{x^2-1}}{(x-1)^2}$ $=\dfrac{x\sqrt{x-1}-\sqrt{(x^2-1)(x+1)}}{(x-1)^2\sqrt{x+1}}$ Trả lời
$y=\sqrt{x+1}.\dfrac{1}{\sqrt{x-1}}$
$→y’=(\dfrac{\sqrt{x+1}}{\sqrt{x-1}})’$
$→y’=\dfrac{(\sqrt{x+1})’\sqrt{x-1}-\sqrt{x+1}(\sqrt{x-1})’}{x-1}$
$→y’=\dfrac{\dfrac{\sqrt{x-1}}{2\sqrt{x+1}}-\dfrac{\sqrt{x+1}}{2\sqrt{x-1}}}{x-1}$
$→y’=\dfrac{\dfrac{-1}{\sqrt{x²-1}}}{x-1}$
$→y’=\dfrac{-1}{\sqrt{x²-1}(x-1)}$
$y=\sqrt{x+1}.\dfrac{1}{\sqrt{x-1}}$
$=\dfrac{\sqrt{x+1}}{\sqrt{x-1}}$
$=\dfrac{\sqrt{x^2-1}}{x-1}$
$y’=\dfrac{(\sqrt{x^2-1})'(x-1)-\sqrt{x^2-1}(x-1)’}{(x-1)^2}$
$=\dfrac{\dfrac{(x^2-1)’}{2\sqrt{x^2-1}}.(x-1)-\sqrt{x^2-1}}{(x-1)^2}$
$=\dfrac{\dfrac{x}{\sqrt{x^2-1}}(x-1)-\sqrt{x^2-1}}{(x-1)^2}$
$=\dfrac{\dfrac{x\sqrt{x-1}}{\sqrt{x+1}}-\sqrt{x^2-1}}{(x-1)^2}$
$=\dfrac{x\sqrt{x-1}-\sqrt{(x^2-1)(x+1)}}{(x-1)^2\sqrt{x+1}}$