Tính đạo hàm của hàm số \(y=\dfrac{\left(1-x\right)^p}{\left(1+x\right)^q}\).
\(y'=\dfrac{\left(1-x\right)^{p-1}\left[\left(p-q\right)x+p+q\right]}{\left(1+x\right)^{q+1}}\) \(y'=\dfrac{-\left(1-x\right)^{p-1}\left[\left(p-q\right)x+p+q\right]}{\left(1+x\right)^{q+1}}\) \(y'=\dfrac{-\left(1-x\right)^{p-1}\left[\left(p+q\right)x+p-q\right]}{\left(1+x\right)^{q+1}}\) \(y'=\dfrac{-\left(1-x\right)^{p-1}\left[\left(p-q\right)x-p+q\right]}{\left(1+x\right)^{q+1}}\) Hướng dẫn giải:\(y=\dfrac{\left(1-x\right)^p}{\left(1+x\right)^q}=\dfrac{u}{v}\) với \(u=\left(1-x\right)^p,v=\left(1+x\right)^q\). Áp dụng quy tắc đạo hàm một thương \(y'=\dfrac{u'v-uv'}{v^2}\).
Ta có \(u'=-p\left(1-x\right)^{p-1},v'=q\left(1+x\right)^q\)
\(u'v-uv'=-p\left(1-x\right)^{p-1}\left(1+x\right)^q-\left(1-x\right)^pq\left(1+x\right)^{q-1}=\left(1-x\right)^{p-1}\left(1+x\right)^q\left[-p\left(1+x\right)-q\left(1-x\right)\right]=-\left(1-x\right)^{p-1}\left(1+x\right)^{q-1}[\left(p-q\right)x+p+q]\)
Từ đó \(y'=\dfrac{-\left(1-x\right)^{p-1}\left(1+x\right)^{q-1}\left[\left(p-q\right)x+p+q\right]}{\left(1+x\right)^{2q}}=\dfrac{-\left(1-x\right)^{p-1}\left[\left(p-q\right)x+p+q\right]}{\left(1+x\right)^{q+1}}\)