Question

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

Answer 1

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)`

Answer 2

`y^((n)) = (-1)^n \ ( (2n)! ) / ( (4^n) \ n!) x^(-(2n+1)/2) \ \ \ n in NN`

Explanation:

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: