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

First, it's recommended to obtain a formula for the #n#th derivative of #cosx#.

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

So we will begin by taking the first derivative:

#dy/dx = -sinx#

Next, the second derivative:

#(d^2y)/(dx^2) = -cosx#

And the third derivative:

#(d^3y)/(dx^3) = sinx#

The fourth:

#(d^4y)/(dx^4) = cosx#

There - we've arrived back at #cosx#. 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 #pi/2#.

What do I mean? Well, #-sin x# is the same thing as #cos(x + pi/2)#. And #-cos x# is the same thing as #cos(x + pi)#.

So there's our formula:

#f^n(x) = cos(x + (pi n)/2)#

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

#f^50(x) = cos(x + 25pi)#

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

#f^50(x) = cos(x + pi)#

Which is the same thing as:

#f^50(x) = -cosx#