Answer
$y^{(157)}=\cos{x}$
Work Step by Step
Firstly calculate the first four derivatives of $y=\sin{x}$.
$y'=\cos{x}$
$y''=-\sin{x}$
$y'''=-\cos{x}$
$y^{(4)}=\sin(x)$
We see that after differentiating four times we get the same function.
So, $y'=y^{(5)}=y^{(9)}=y^{(4\times n+1)}$ where $n$ is any natural number.
Similarly,
$y''=y^{(4\times n+2)}$, $y'''=y^{(4\times n+3)}$ and $y^{(4)}=y^{(4\times n+4)}=y^{(4\times (n+1))}$, where $n$ is a natural number.
Now, divide $157$ by $4$.
we get, $157=4\times39+1$
So, we get $y^{(157)}=y^{(4\times39+1)}=y^{(4\times n+1)}$ where $n=39$.
Now subtitute $y^{(4\times n+1)}=y'$.
We get, $y^{(157)}=y'$
Now substitute $y'=\cos{x}$.
We get, $y^{(157)}=\cos{x}$
