Tính đạo hàm của hàm số \(y=\dfrac{x}{\left(1-x^2\right)\left(1+x\right)^3}\).
\(y'=\dfrac{4x^2-x+1}{\left(1-x\right)^3\left(1+x\right)^4}\).\(y'=\dfrac{4x^2+3x+1}{\left(1-x^2\right)^2\left(1+x\right)^3}\). \(y'=\dfrac{4x^2-3x+1}{\left(1-x\right)^2\left(1+x\right)^5}\).\(y'=\dfrac{4x^2+x+1}{\left(1-x\right)^3\left(1+x\right)^4}\).Hướng dẫn giải:\(y=\dfrac{x}{\left(1-x^2\right)\left(1+x\right)^3}=\dfrac{u}{v}\) với \(u=x,v=\left(1-x^2\right)\left(1+x\right)^3\). Theo quy tắc đạo hàm một thương thì \(y'=\dfrac{u'v-uv'}{v^2}\) trong đó
\(u'=1,v'=-2x\left(1+x\right)^3+\left(1-x^2\right)3\left(1+x\right)^2=\left(1+x\right)^3\left[-2x+3\left(1-x\right)\right]=\left(1+x\right)^3\left(-5x+3\right)\)
\(u'v-uv'=1.\left(1-x^2\right)\left(1+x\right)^3-x.\left(1+x\right)^3\left(-5x+3\right)=\left(1+x\right)^3\left[\left(1-x^2\right)-x\left(-5x+3\right)\right]=\left(1+x\right)^3\left(4x^2-3x+1\right)\)
\(y'=\dfrac{\left(1+x\right)^3\left(4x^2-3x+1\right)}{\left(1-x^2\right)^2\left(1+x\right)^6}=\dfrac{4x^2-3x+1}{\left(1-x^2\right)^2\left(1+x\right)^3}=\dfrac{4x^2-3x+1}{\left(1-x\right)^2\left(1+x\right)^5}\)
