# How do you find the 50th derivative of y=cos(x) ?

Aug 1, 2014

First, it's recommended to obtain a formula for the $n$th derivative of $\cos x$.

To do this, usually it is needed to continually differentiate until you notice a pattern.

So we will begin by taking the first derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin x$

Next, the second derivative:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \cos x$

And the third derivative:

$\frac{{d}^{3} y}{{\mathrm{dx}}^{3}} = \sin x$

The fourth:

$\frac{{d}^{4} y}{{\mathrm{dx}}^{4}} = \cos x$

There - we've arrived back at $\cos x$. Since this was our original function, differentiating again will just give us the first derivative, and so on. So, we can deduce that the $n$th derivative is periodic.

Now the problem is putting this pattern into a formula. At first it might look like there's no mathematically explainable pattern - we have a negative, then a negative, then a positive, then a positive, meanwhile flipping from sine to cosine - but when you graph these successive functions, it's easy to see that each graph is the previous derivative, but shifted to the left by $\frac{\pi}{2}$.

What do I mean? Well, $- \sin x$ is the same thing as $\cos \left(x + \frac{\pi}{2}\right)$. And $- \cos x$ is the same thing as $\cos \left(x + \pi\right)$.

So there's our formula:

${f}^{n} \left(x\right) = \cos \left(x + \frac{\pi n}{2}\right)$

Now, if we substitute $n = 50$, we obtain:

${f}^{50} \left(x\right) = \cos \left(x + 25 \pi\right)$

Since cosine itself is periodic, we can divide the $25$ by $2$ and leave the remainder next to the $\pi$:

${f}^{50} \left(x\right) = \cos \left(x + \pi\right)$

Which is the same thing as:

${f}^{50} \left(x\right) = - \cos x$