Tính đạo hàm của hàm số \(y=\left[\frac{1-x^2}{2}\sin x-\frac{\left(1+x\right)^2}{2}\cos x\right]e^{-x}\).
\(x^2\cos x.e^{-x}\) \(x\sin x.e^{-x}\) \(x^2\sin x.e^{-x}\) \(x\cos x.e^{-x}\) Hướng dẫn giải:Đặt \(u=\dfrac{1-x^2}{2}\sin x-\dfrac{\left(1+x\right)^2}{2}\cos x\) và \(v=e^{-x}\) thì \(y=uv,y'=u'v+uv'\).
Tính \(u'\): \(u'=\left(-x\sin x+\dfrac{1-x^2}{2}\cos x\right)-\left(\left(1+x\right)\cos x-\dfrac{\left(1+x\right)^2\sin x}{2}\right)\)
\(=\left(-x+\dfrac{\left(1+x\right)^2}{2}\right)\sin x+\left(\dfrac{1-x^2}{2}-\left(1+x\right)\right)\cos x\)
\(=\dfrac{x^2+1}{2}\sin x-\dfrac{x^2+2x+1}{2}\cos x=\dfrac{x^2+1}{2}\sin x-\dfrac{\left(1+x\right)^2}{2}\cos x\)
Tính \(v'\): \(v'=-e^{-x}\)
Do đó \(\left\{{}\begin{matrix}u'v=\left(\dfrac{x^2+1}{2}\sin x-\dfrac{\left(1+x\right)^2}{2}\cos x\right)e^{-x}\\uv'=-\left(\dfrac{1-x^2}{2}\sin x-\dfrac{\left(1+x\right)^2}{2}\cos x\right)e^{-x}\end{matrix}\right.\) \(\Rightarrow y'=u'v+uv'=\left(\dfrac{x^2+1}{2}-\dfrac{1-x^2}{2}\right)\sin x.e^{-x}=x^2\sin xe^{-x}\)
Đáp số: \(x^2\sin x.e^{-x}\)