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

Aug 23, 2017

I hope I'm interpreting this correctly.

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

$y = {x}^{- \frac{1}{2}} \text{ "=>" } y \left(\textcolor{b l u e}{x}\right) = {\textcolor{b l u e}{x}}^{- \frac{1}{2}}$

Then,

$y \left(\textcolor{b l u e}{n}\right) = {\textcolor{b l u e}{n}}^{- \frac{1}{2}}$

Aug 23, 2017

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

#### Explanation:

We seek the ${n}^{t h}$ derivative:

${y}^{\left(n\right)} = \frac{{d}^{n}}{{\mathrm{dx}}^{n}} {x}^{- \frac{1}{2}}$

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

$y ' \setminus \setminus \setminus = \left(- \frac{1}{2}\right) {x}^{- \frac{1}{2} - 1}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{1}{2} {x}^{- \frac{3}{2}}$

And differentiating again we get:

 y'' \ \ = (-1/2)(-3/2)x^(-3/2-1 )
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{3}{2} ^ 2 {x}^{- \frac{5}{2}}$

And differentiating again we get:

${y}^{\left(3\right)} = \frac{3}{2} ^ 2 \left(- \frac{5}{2}\right) {x}^{- \frac{7}{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{3.5}{2} ^ 3 {x}^{- \frac{7}{2}}$

And differentiating again we get:

${y}^{\left(4\right)} = - \frac{3.5}{2} ^ 3 \left(- \frac{7}{2}\right) {x}^{- \frac{9}{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{3.5 .7}{2} ^ 4 {x}^{- \frac{9}{2}}$

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

${y}^{\left(n\right)} = {\left(- 1\right)}^{n} \setminus \frac{1.3 .5 . . \left(2 n - 1\right)}{{2}^{n}} {x}^{- \frac{2 n + 1}{2}}$

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

$1.3 .5 \ldots \left(2 n - 1\right) = \frac{1.2 .3 .4 .5 \ldots \left(2 n - 1\right) \left(2 n\right)}{2.4 .6 \ldots \left(2 n\right)}$
$\text{ } = \frac{1.2 .3 .4 .5 \ldots \left(2 n - 1\right) \left(2 n\right)}{\left(1.2\right) \left(2.2\right) \left(3.2\right) \ldots \left(n .2\right)}$
$\text{ } = \frac{1.2 .3 .4 .5 \ldots \left(2 n - 1\right) \left(2 n\right)}{{2}^{n} \left(1.2 .3 \ldots n\right)}$
 " " = ( (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: