Ngân hàng bài tập
SS
Cho hàm số $y=\sin^2x$. Tính $y^{\left(2018\right)}\left(\pi\right)$.
 | $y^{\left(2018\right)}\left(\pi\right)=2^{2017}$ |
 | $y^{\left(2018\right)}\left(\pi\right)=2^{2018}$ |
 | $y^{\left(2018\right)}\left(\pi\right)=-2^{2017}$ |
 | $y^{\left(2018\right)}\left(\pi\right)=-2^{2018}$ |
1 lời giải
Chọn phương án A.
- $y'=\sin2x$
- $y''=2\cos2x=2\sin\left(2x+\dfrac{\pi}{2}\right)$
- $y'''=-2^2.\sin2x=2^2.\sin\left(2x+\pi\right)$
......
Dự đoán: $y^{\left(n\right)}=2^{n-1}\sin\left[2x+\dfrac{\left(n-1\right)\pi}{2}\right]$.
$\begin{aligned}
\Rightarrow y^{\left(2018\right)}&=2^{2017}.\sin\left(2\pi+\dfrac{2017\pi}{2}\right)\\
&=2^{2017}.\sin\left(1010\pi+\dfrac{\pi}{2}\right)\\
&=2^{2017}.
\end{aligned}$