Question

What is the formula for the `n`th derivative of `f(x) = x^(1/2)` ?

Answer 1

`f^((n))(x) = (-1)^(n-1) ((2n-2)!) / (2^(2n-1) (n-1)!) x^((1-2n)/2)`

Explanation:

Let us look at the first few derivatives to see what's happening:

`f^((0))(x) = x^(1/2)`

`f^((1))(x) = 1/2 x^(-1/2)`

`f^((2))(x) = -1/4 x^(-3/2)`

`f^((3))(x) = 3/8 x^(-5/2)`

`f^((4))(x) = -15/16 x^(-7/2)`

`f^((5))(x) = 105/32 x^(-9/2)`

`f^((6))(x) = -945/64 x^(-11/2)`

The coefficient is a product of odd numbers divided by a power of `2`.

Note that:

`1 = (1 * 2)/2 = (2!) / (2^1 * 1!)`

`1 * 3 = (1 * 2 * 3 * 4) / (2 * 4) = (4!) / (2^2 * 2!)`

`1 * 3 * 5 = (1 * 2 * 3 * 4 * 5 * 6) / (2 * 4 * 6) = (6!) / (2^3 * 3!)`

etc.

So we can write:

`f^((2))(x) = (-1)^(2-1) (2!) / (2^3 * 1!) x^(-3/2)`

`f^((3))(x) = (-1)^(3-1) (4!) / (2^5 * 2!) x^(-5/2)`

`f^((4))(x) = (-1)^(4-1) (6!) / (2^7 * 3!) x^(-7/2)`

`f^((5))(x) = (-1)^(5-1) (8!) / (2^9 * 4!) x^(-9/2)`

So it looks like a valid formula for `n > 1` would be:

`f^((n))(x) = (-1)^(n-1) ((2n-2)!) / (2^(2n-1) (n-1)!) x^((1-2n)/2)`

If `n = 1` then it gives us:

`f^((1))(x) = (-1)^(1-1) ((2-2)!) / (2^(2-1) (1-1)!) x^((1-2)/2)`

`color(white)(f^((1))(x)) = 1/2 x^(-1/2)`

which is correct too.

So this formula seems to work for all `n >= 1`