Question

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

Answer 1

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`