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

Nov 11, 2016

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

#### Explanation:

Let us look at the first few derivatives to see what's happening:

${f}^{\left(0\right)} \left(x\right) = {x}^{\frac{1}{2}}$

${f}^{\left(1\right)} \left(x\right) = \frac{1}{2} {x}^{- \frac{1}{2}}$

${f}^{\left(2\right)} \left(x\right) = - \frac{1}{4} {x}^{- \frac{3}{2}}$

${f}^{\left(3\right)} \left(x\right) = \frac{3}{8} {x}^{- \frac{5}{2}}$

${f}^{\left(4\right)} \left(x\right) = - \frac{15}{16} {x}^{- \frac{7}{2}}$

${f}^{\left(5\right)} \left(x\right) = \frac{105}{32} {x}^{- \frac{9}{2}}$

${f}^{\left(6\right)} \left(x\right) = - \frac{945}{64} {x}^{- \frac{11}{2}}$

The coefficient is a product of odd numbers divided by a power of $2$.

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 $n > 1$ would be:

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

If $n = 1$ then it gives us:

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

$\textcolor{w h i t e}{{f}^{\left(1\right)} \left(x\right)} = \frac{1}{2} {x}^{- \frac{1}{2}}$

which is correct too.

So this formula seems to work for all $n \ge 1$