Tính đạo hàm của hàm số \(y=\dfrac{x}{\left(1-x\right)^2\left(1+x\right)^3}\).
\(y'=\dfrac{4x^2-x+1}{\left(1-x\right)^3\left(1+x\right)}\).\(y'=\dfrac{4x^2-x+1}{\left(1-x\right)\left(1+x\right)^4}\).\(y'=\dfrac{4x^2-x+1}{\left(1-x\right)^3\left(1+x\right)^4}\).\(y'=\dfrac{4x^2-x+1}{\left(1-x\right)^2\left(1+x\right)^4}\).Hướng dẫn giải:\(y=\dfrac{x}{\left(1-x\right)^2\left(1+x\right)^3}=\dfrac{u}{v}\), trong đó \(u=x,v=\left(1-x\right)^2\left(1+x\right)^3\) . Theo quy tắc đạo hàm một thương: \(y'=\dfrac{u'v-uv'}{v^2}\).
Ta có \(u'=1,v'=-2\left(1-x\right)\left(1+x\right)^3+\left(1-x\right)^23\left(1+x\right)^2=\left(1-x\right)\left(1+x\right)^2\left[-2\left(1+x\right)+3\left(1-x\right)\right]=\left(1-x\right)\left(1+x\right)^2\left(-5x+1\right)\).
\(u'v-uv'=1.\left(1-x\right)^2\left(1+x\right)^3-x\left(1-x\right)\left(1+x\right)^2\left(-5x+1\right)=\left(1-x\right)\left(1+x\right)^2\left[\left(1-x^2\right)-x\left(-5x+1\right)\right]=\left(1-x\right)\left(1+x\right)^2\left(4x^2-x+1\right)\)
\(y'=\dfrac{\left(1-x\right)\left(1+x\right)^2\left(4x^2-x+1\right)}{\left(1-x\right)^4\left(1+x\right)^6}=\dfrac{4x^2-x+1}{\left(1-x\right)^3\left(1+x\right)^4}\)
