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

##### 1 Answer

#### 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

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

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

If

#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