Question

Tính y'

a) \(y=\sqrt{1-2x}+\sqrt{1+2x}\)

b) \(y=\dfrac{x+1}{\sqrt{x^2+1}}\)

Answer 1

a. \(y'=\dfrac{-2}{2\sqrt{1-2x}}+\dfrac{2}{2\sqrt{1+2x}}=\dfrac{1}{\sqrt{1+2x}}-\dfrac{1}{\sqrt{1-2x}}\)

b. \(y'=\dfrac{\sqrt{x^2+1}-\dfrac{x\left(x+1\right)}{\sqrt{x^2+1}}}{x^2+1}=\dfrac{x^2+1-\left(x^2+x\right)}{\left(x^2+1\right)\sqrt{x^2+1}}=\dfrac{1-x}{\left(x^2+1\right)\sqrt{x^2+1}}\)