Answer
$f'(x)=-\sin{x}$
$f''(x)=-\cos{x}$
$f'''(x)=\sin{x}$
$f^{(4)}(x)=\cos{x}$
$f^{(5)}(x)=-\sin{x}$
$f^{(8)}(x)=\cos{x}$
$f^{(37)}(x)=-\sin{x}$
Work Step by Step
Firstly find the first derivative of $f (x) = cos x$.
That is, $f'(x)=(\cos{x})'=-\sin{x}$ or $f'(x)=-1\times \sin{x}$
Now again differentiate using the product rule to find the second derivative.
$f''(x)=(-1\times \sin{x})'$
$\hspace{45px}=(-1)'\sin{x}+(\sin{x})'(-1)$
$\hspace{45px}=0\times\sin{x}+\cos{x}(-1)$
$\hspace{45px}=-\cos{x}$
Now again differentiate using the product rule to find the third derivative.
$f'''(x)=(-1\times \cos{x})'$
$\hspace{45px}=(-1)'\cos{x}+(\cos{x})'(-1)$
$\hspace{45px}=0\times\cos{x}+(-\sin{x})(-1)$
$\hspace{45px}=\sin{x}$
Now again differentiate to find the fourth derivative.
$f^{(4)}(x)=( \sin{x})'=\cos{x}$
Now again differentiate to find the fifth derivative.
$f^{(5)}(x)=( \cos{x})'=-\sin{x}$
We see that after derivating four times, we get the same value.
That is, $f^{(n+4)}(x)=f^{(n)}(x)$
Now, substitute $n=4$.
$f^{(8)}(x)=f^{(4)}(x)$
Now, substitute $f^{(4)}(x)=\cos{x}$.
We get, $f^{(8)}(x)=\cos{x}$.
Similarly,
$f^{(37)}(x)=f^{(33)}(x)=f^{(29)}(x)=f^{(25)}(x)=f^{(21)}(x)=f^{(17)}(x)=f^{(13)}(x)=f^{(9)}(x)=f^{(5)}(x)$
Hence, we get $f^{(37)}(x)=f^{(5)}(x)$.
Now, substitute $f^{(5)}(x)=-\sin{x}$.
We get, $f^{(37)}(x)=-\sin{x}$.
