# Half derivative of x

The derivative of x is easy, it’s simply 1; the second derivative of x is the derivative of 1, which is 0; but what on earth is the half derivative of x??

When I came across this craziness, I was honestly shocked. I’d never thought of anything like that before, so I just couldn’t wrap my head around it. But don’t worry, it’s not as bad as it might seem.

Let’s start by giving a look at the result of first, second, third, fourth derivative of x to the first, second, third and fourth power to see if we can find any pattern:

$\frac{\mathrm{d}&space;}{\mathrm{d}&space;x}x=1$ $\frac{\mathrm{d^2}&space;}{\mathrm{d}&space;x^2}\,x^2=2$ $\frac{\mathrm{d^3}&space;}{\mathrm{d}&space;x^3}\,x^3=6$ $\frac{\mathrm{d^4}&space;}{\mathrm{d}&space;x^4}\,x^4=24$

There actually is a pattern! Take a look at this:

$\frac{\mathrm{d}&space;}{\mathrm{d}&space;x}\,x=1$ $\frac{\mathrm{d^2}&space;}{\mathrm{d}&space;x^2}\,x^2=2\cdot&space;1=2$ $\frac{\mathrm{d^3}&space;}{\mathrm{d}&space;x^3}\,x^3=3\cdot&space;2\cdot&space;1=6$ $\frac{\mathrm{d^4}&space;}{\mathrm{d}&space;x^4}\,x^4=4\cdot&space;3\cdot&space;2\cdot&space;1=24$

This means the fifth derivative of x^5 is 5 · 4 · 3 · 2 · 1 = 120 and so on. This operation consisting of multiplying all numbers from n to 1 is called factorial and is represented by the symbol ‘!’.

$0!=1$ $1!=1$ $2!=2\cdot&space;1=2$ $3!=3\cdot&space;2\cdot&space;1=6$ $4!=4\cdot&space;3\cdot&space;2\cdot&space;1=24$

Therefore,

$\frac{\mathrm{d}^n&space;}{\mathrm{d}&space;x^n}\,x^n=n!$

In our case, however, we have the half derivative of x, i.e.

$\frac{\mathrm{d}^n&space;}{\mathrm{d}&space;x^n}\,x^{2n}$

Let’s do some examples:

$\frac{\mathrm{d}&space;}{\mathrm{d}&space;x}\,x^2=2x$ $\frac{\mathrm{d^2}&space;}{\mathrm{d}&space;x^2}\,x^4=12x^2$ $\frac{\mathrm{d^3}&space;}{\mathrm{d}&space;x^3}\,x^6=120x^3$ $\frac{\mathrm{d^4}&space;}{\mathrm{d}&space;x^4}\,x^8=1680x^4$

From this we know for sure that

$\frac{\mathrm{d^{\frac{1}{2}}}&space;}{\mathrm{d}&space;x^{\frac{1}{2}}}\,x=kx^\frac{1}{2}$ $\frac{\mathrm{d}^n&space;}{\mathrm{d}&space;x^n}\,x^{2n}=kx^n$

The exponent of the x is equal to the order of the derivative. Now we have to figure out what k is.

Let’s decompose the coefficients of the x to see if there’s any pattern:

$2=2$ $12=2^2\cdot&space;3$ $120=2^3\cdot&space;3\cdot&space;5$ $1680=2^4\cdot&space;3\cdot&space;5\cdot&space;7$

As you can see, the exponent of the 2 is equal to the order n of the derivative, therefore we have

$\frac{\mathrm{d}&space;^n}{\mathrm{d}&space;x^n}\,x^&space;{2n}=2^njx^n$

Now we need to find j.

Notice that the greatest number in each multiplication has to do with n; 1680 is the coefficient of x^4 and, as we’ve just seen, the greatest term is 7, which is given by 4+3, where 4 is n. The same is for 120, where the highest term is 5, i.e. 3+2, where n is 3, and so on. Every time we sum n to n-1, so we know that the greatest term is equal to n+n-1=2n-1. We can also see that before 7 is 5, which is 7-2, or 2n-1-2, so 2n-3, then we have 3, which is 2n-5, ans as last is 1, i.e. 2n-7.

$\frac{\mathrm{d}&space;^4}{\mathrm{d}&space;x^4}\,x^&space;{8}=2^4(8-1)(8-3)(8-5)(8-7)x^4=2^4\cdot&space;7\cdot&space;5\cdot&space;3\cdot&space;1x^4$

1 · 3 · 5 · 7 looks a lot like 1 · 2 · 3 · 4 · 5 · 6 · 7, which is 7!; the only difference is that it’s all divided by 2 · 4 · 6:

$1\cdot&space;3\cdot&space;5\cdot&space;7=\frac{1\cdot&space;2\cdot&space;3\cdot&space;4\cdot&space;5\cdot&space;6\cdot&space;7}{2\cdot&space;4\cdot&space;6}=\frac{7!}{2^3(1\cdot&space;2\cdot&space;3)}=\frac{7!}{2^3\cdot&space;3!}$

Therefore,

$\frac{\mathrm{d}&space;^4}{\mathrm{d}&space;x^4}\,x^8=2^4\frac{7!}{2^3\cdot&space;3!}x^4=\frac{2\cdot&space;7!}{3!}\,x^4=1680x^4$

We get the correct result! At this point we notice that 3 is equal to n-1 and 7 is equal to 2n-1, so we can generalize this expression for any n:

$\frac{\mathrm{d}&space;^n}{\mathrm{d}&space;x^n}\,x^{2n}=\frac{2^n\cdot&space;(2n-1)!}{2^{n-1}\cdot&space;(n-1)!}x^{2n}=\frac{2(2n-1)!}{(n-1)!}x^{2n}$

This is the formula for the n-th derivative of x to the 2n:

$\frac{\mathrm{d}&space;^n}{\mathrm{d}&space;x^n}\,x^{2n}=\frac{2(2n-1)!}{(n-1)!}\,x^n$

Now we can find the half derivative of x:

$\frac{\mathrm{d}&space;^\frac{1}{2}}{\mathrm{d}&space;x^\frac{1}{2}}\,x=\frac{2(2\frac{1}{2}-1)!}{(\frac{1}{2}-1)!}\,x^\frac{1}{2}=\frac{2}{\left&space;(&space;-\frac{1}{2}&space;\right&space;)!}x^{\frac{1}{2}}$

The factorial function is limited to positive integers, such as 0, 1, 2, 3 and so on. For negative and/or non-integer numbers, the gamma function is used, which is actually the definition of (n-1)! using integrals, but I will talk about that in another post. To find the factorial of -1/2, just type (-1/2)! on Google or on a scientific calculator and this amazing result will show up:

$\left&space;(-\frac{1}{2}&space;\right&space;)!=\sqrt{\pi}$

Finally, we find that

$\frac{\mathrm{d}&space;^\frac{1}{2}}{\mathrm{d}&space;x^\frac{1}{2}}\,x=\frac{2}{\sqrt{\pi}}\sqrt&space;x$

That’s it!

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