giuap mk đi

$\displaystyle f( x) =\sin( \pi x)$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow f'( x) =\pi \cos \pi x=\sin\left( \pi x+\frac{\pi }{2}\right)\\
\Rightarrow f''( x) =-\pi ^{2}\sin \pi x=\pi ^{2}\sin( \pi x+\pi )\\
\Rightarrow f'''( x) =-\pi ^{3}\cos \pi x=\pi ^{3}\sin\left( \pi x+\frac{3\pi }{2}\right)
\end{array}$
Dự đoán $\displaystyle f^{( n)}( x) =\pi ^{n}\sin\left( \pi x+\frac{n\pi }{2}\right) \ ,n\in N$
Ta chứng minh bằng phương pháp quy nạp
Với $\displaystyle n=1\ \Rightarrow f'( x) =\pi \cos( \pi x) =\pi \sin\left( \pi x+\frac{\pi }{2}\right) \ $(đúng)
Giả sử công thức đúng với $\displaystyle n=k\ ( k\in N,k >1)$
$\displaystyle \Rightarrow f^{( k)}( x) =\pi ^{k}\sin\left( \pi x+\frac{k\pi }{2}\right)$
Ta phải chứng minh công thức đúng với $\displaystyle n=k+1$
hay $\displaystyle f^{( k+1)}( x) =\pi ^{k+1}\sin\left( \pi x+\frac{( k+1) \pi }{2}\right)$
Có $\displaystyle f^{( k+1)}( x) =\left[ f^{( k)}( x)\right] '=\left[ \pi ^{k}\sin\left( \pi x+\frac{k\pi }{2}\right)\right] '=\pi ^{k+1}\cos\left( \pi x+\frac{k\pi }{2}\right)$
$\displaystyle =\pi ^{k+1}\cos\left( -\pi x-\frac{k\pi }{2}\right) =\pi ^{k+1}\sin\left(\frac{\pi }{2} +\pi x+\frac{k\pi }{2}\right) =\pi ^{k+1}\sin\left( \pi x+\frac{( k+1) \pi }{2}\right) \ ( đúng)$
Vậy $\displaystyle f^{( n)}( x) =\pi ^{n}\sin\left( \pi x+\frac{n\pi }{2}\right) \ $
$\displaystyle \Rightarrow f^{( 2023)}( x) =\pi ^{2023} .\sin\left( \pi x+\frac{2023\pi }{2}\right) =\pi ^{2023} .-\sin\left( \pi x+\frac{\pi }{2}\right) =-\pi ^{2023} .\cos( \pi x)$