What is the #n^(th)# derivative of #x^(-1/2) #?

I hope I'm interpreting this correctly.

The function #y#, as a function of #x#, can also be notated as #y(x)#. Then, we could have:

#y=x^(-1/2)" "=>" "y(color(blue)x)=color(blue)x^(-1/2)#

Then,

#y(color(blue)n)=color(blue)n^(-1/2)#

We seek the #n^(th)# derivative:

# y^((n)) = (d^n)/(dx^n) x^(-1/2) #

This can readily be formed by repeated application of the power rule, viz:

# y' \ \ \ = (-1/2)x^(-1/2-1) #
# \ \ \ \ \ \ \ = -1/2x^(-3/2) #

And differentiating again we get:

# y'' \ \ = (-1/2)(-3/2)x^(-3/2-1 #)
# \ \ \ \ \ \ \ = 3/2^2 x^(-5/2) #

And differentiating again we get:

# y^((3)) = 3/2^2 (-5/2)x^(-7/2) #
# \ \ \ \ \ \ \ = -(3.5)/2^3 x^(-7/2) #

And differentiating again we get:

# y^((4)) = -(3.5)/2^3 (-7/2)x^(-9/2) #
# \ \ \ \ \ \ \ = (3.5.7)/2^4 x^(-9/2) #

And we can see a clear pattern forming, and that:

# y^((n)) = (-1)^n \ (1.3.5..(2n-1))/(2^n) x^(-(2n+1)/2)#

We can simplify this further as we not the product if the odd consecutive integers can be written as:

# 1.3.5...(2n-1) = ( 1.2.3.4.5 ... (2n-1)(2n) ) / ( 2.4.6...(2n) ) #
# " " = ( 1.2.3.4.5 ... (2n-1)(2n) ) / ( (1.2)(2.2)(3.2)...(n.2)) #
# " " = ( 1.2.3.4.5 ... (2n-1)(2n) ) / ( 2^n(1.2.3...n)) #
# " " = ( (2n)! ) / ( 2^n \ n!) #

Hence, we have:

# y^((n)) = (-1)^n \ (( (2n)! ) / ( 2^n \ n!))/(2^n) x^(-(2n+1)/2)#

# " " = (-1)^n \ ( (2n)! ) / ( (2^n)^2 \ n!) x^(-(2n+1)/2)#

# " " = (-1)^n \ ( (2n)! ) / ( (4^n) \ n!) x^(-(2n+1)/2)#

A formal proof can readily be established via induction, should it be required: