$99$th derivative of $\sin x$ Can someone help me calculate the $99$th derivative of $\sin(x)$?
Calculate $f^{(99)}(x) $ for the function $f(x) = \sin(x) $
 A: We have
$$\left(\sin x\right)^{(k)}=\sin\left( x+k\frac\pi2\right)$$ 
and recall that the $\sin$ function is $2\pi$ periodic.
A: Note that given $f(x)=\sin x,$
\begin{align}
& f'(x)=\cos x\\
& f''(x)=-\sin x\\
& f'''(x)=-\cos x\\
& f^{(4)}(x)=\sin x
\end{align}
Why is this important? Well, note that after taking the derivative of $\sin x$ four times, we are back at $\sin x$. So, we can conclude that for $f(x)=\sin x,$
\begin{align}
 f(x)=f^{(4)}(x)=f^{(8)}(x)=\ldots=f^{(96)}(x)=\sin x
\end{align}
So given that $f^{(96)}(x)=\sin x$, then $f^{(97)}(x)=\cos x, f^{(98)}(x)=-\sin x, f^{(99)}(x)=-\cos x.$
If you are familiar with integration, you also can note that $f(x)=\sin x=f^{(100)}(x)$, and so $f^{(99)}(x)=\int\sin x=-\cos x$.
A: Do you know modular arithmetic? Notice from the first few derivatives,


*

*$f(x)=\sin(x)$

*$f'(x)=\cos(x)$

*$f''(x)=-\sin(x)$

*$f'''(x)=-\cos(x)$

*$f^{(4)}(x)=\sin(x)$


Therefore, we have that
$f^{(n)}(x) = \left\{ \begin{array} \. \sin(x), & n \equiv 0 (\mod 4)\\
\cos(x), & n \equiv 1 (\mod 4)\\
 -\sin(x), & n \equiv 2 (\mod 4)\\
- \cos(x), & n \equiv 3 (\mod 4).
\end{array} \right.$
So, notice that $99 \div 4=24$ with remainder $3$. Therefore, $f^{(99)}(x)=-\cos(x)$.
A: Notice if $f(x) = \sin x$
$f'(x) = \cos x = \sin( x + \frac{\pi}{2}) $
$f''(x) = - \sin x = \sin( x + \pi) = \sin( x + 2 (\frac{\pi}{2})) $
$f'''(x) = - \cos x = \sin(x + 3( \frac{\pi}{2} ))$
$f''''(x) = \sin x $
Hence, we can say that 
$$ f^{(n)} (x) = \sin \left( x + n \cdot\frac{\pi}{2} \right) $$
A: ${{d^{4n}}\over{dx^{4n}}}(\sin x)=\sin{x}$
${{d^{4n+1}}\over{dx^{4n+1}}}(\sin x)=\cos{x}$
${{d^{4n+2}}\over{dx^{4n+2}}}(\sin x)=-\sin{x}$
${{d^{4n+3}}\over{dx^{4n+3}}}(\sin x)=-\cos{x}$
Pick the one that you think might be useful :-)
