# What is the derivative of x^(3/2)?

##### 1 Answer
May 21, 2017

$\frac{3}{2} {x}^{\frac{1}{2}}$

#### Explanation:

If you have already learnt the concepts of differentiating, skip to the solutions instead.

Differentiation of power functions is found as $\textcolor{red}{f ' \left(x\right)} = n {x}^{n - 1}$.
By differentiating a function, you are decreasing its power/exponent by 1.

Imagine you are given a cube with its corner lengths being $x c m$. This cube has 3 dimensions, with is its length, width and height.

The volume of this cube will be $x . x . x = {x}^{3} c {m}^{3}$

Thus, let the volume be $V \left(x\right) = {x}^{3}$. By differentiating the volume of the cube, we are actually reducing its dimension from 3 to 2.

The result is $V ' \left(x\right) = 3 {x}^{3 - 1} c {m}^{3 - 1}$

$V ' \left(x\right) = 3 {x}^{2} c {m}^{2}$

Notice we differentiate both the function and its unit.

Now, its power has decreased to 2, and its unit is now cm squared instead of cm cubed.

By differentiating, the cube has been reduced from having 3 dimensions to just 2 dimensions.

The cube has been reduced to just a surface of the cube (or plane) ie changed from cube to a square.

If we differentiate again, $V ' ' \left(x\right) = 3 \cdot 2 {x}^{2 - 1} c {m}^{2 - 1}$

$V ' ' \left(x\right) = 6 x c {m}^{1}$

Its power is reduced to 1, and reduced to just a line from a square.

So what happens when we differentiate $V ' ' \left(x\right)$?

$V ' ' ' \left(x\right) = 6 \cdot 1 {x}^{1 - 0} c {m}^{1 - 1}$
$V ' ' ' \left(x\right) = 0 c {m}^{0}$

Which is basically nothing (zero).

Try practicing differentiating the volume of a sphere, $V = \frac{4}{3} \pi {r}^{3}$ and then its surface area.

Solution

$\frac{d}{\mathrm{dx}} {x}^{\frac{3}{2}} = \left(\frac{3}{2}\right) {x}^{\frac{3}{2} - 1}$

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

Differentiating rational powers of functions ie ${x}^{4}$ or ${x}^{5}$, we can picture reducing its dimension by 1.

Can you imagine what happens by differentiating functions with irrational powers ie ${x}^{\frac{2}{5}}$?