Tính đạo hàm của hàm số \(y=\dfrac{\left(2-x^2\right)\left(3-x^3\right)}{\left(1-x\right)^2}\)
\(y'=\dfrac{-3x^5+5x^4+2x^3-6x^2-6x+12}{\left(1-x\right)^4}\).\(y'=\dfrac{-3x^5+5x^4+2x^3-6x^2-6x+12}{\left(1-x\right)^3}\).\(y'=\dfrac{-3x^5-5x^4+2x^3+6x^2-6x+12}{\left(1-x\right)^3}\).\(y'=\dfrac{-3x^5+x^4+2x^3-3x^2-6x+12}{\left(1-x\right)^3}\).Hướng dẫn giải:\(y=\dfrac{\left(2-x^2\right)\left(3-x^3\right)}{\left(1-x\right)^2}=\dfrac{x^5-2x^3-3x^2+6}{\left(1-x\right)^2}=\dfrac{u}{v}\) với \(u=x^5-2x^3-3x^2+6,v=\left(1-x\right)^2\). Ta có \(y'=\dfrac{u'v-uv'}{v^2}\).
\(u'=5x^4-6x^2-6x,v'=-2\left(1-x\right)\) ; \(v^2=\left(1-x\right)^4\)
\(u'v-uv'=\left(5x^4-6x^2-6x\right)\left(1-x\right)^2+2\left(x^5-2x^3-3x^2+6\right)\left(1-x\right)=\left(1-x\right)\left[\left(5x^4-6x^2-6x\right)\left(1-x\right)+2\left(x^5-2x^3-3x^2+6\right)\right]\)
\(=\left(1-x\right)\left(-3x^5+5x^4+2x^3-6x^2-6x+12\right)\) . Do đó \(y'=\dfrac{-3x^5+5x^4+2x^3-6x^2-6x+12}{\left(1-x\right)^3}\)
