Rational Exponents

You can take the deriative of functions which use rational exponents i.e. fractional exponents in the form

\[ y = x^{\frac{m}{n}} \]

You can do this by simply applying the \( nx^{n-1} \) rule. For example,

\[ \frac{d}{dx}x^{\frac{1}{2}} \]

\[ = \frac{1}{2}x^{-\frac{1}{2}} \]

Proof

We can prove this works by considering the general case for rational exponents

\[ y = x^{\frac{m}{n}} \]

and then using the chain rule and other derivative rules to prove that taking the derivative of this function is the same as the \( nx^{n-1} \) rule.

First, if we can’t differentiate the \( \frac{m}{n} \) directly we can instead raise both sides of the function to power of \( n \) to remove fractional component.

\[ y^n = (x^{\frac{m}{n}})^n \]

We then multiply \( \frac{m}{n} \) and \( n \) to get \( m \).

\[ y^n = x^m \]

These are now both regular exponents so we can try and implicitly find the derivative of \( y \) by taking the derivative of boths sides of the function.

\[ \frac{d}{dx}y^n = \frac{d}{dx}x^m \]

Because \( y \) is itself a function that is raised to \( n \) we apply the chain rule to it and get \( ny^{n-1} \) (the outer function) multiplied by the derivative of \( y \) (the inner function). This value \( \frac{d}{dx}y \) is what we currently don’t know and are trying to find! Because that was the original derivative we were trying to take at the very beginning but didn’t know how to.

\[ ny^{n-1}\frac{d}{dx}y = mx^{m-1} \]

By dividing by \( ny^{n-1} \) we can isolate \( \frac{d}{dx}y \) and get

\[ \frac{d}{dx}y = \frac{m}{n}\frac{x^{m-1}}{y^{n-1}} \]

We can then subsitute the value of \(y \) as \( y = x^{\frac{m}{n}} \) into this equation as it was originally defined in terms of \( x \) to remove it completely.

\[ \frac{d}{dx}y = \frac{m}{n}\frac{x^{m-1}}{(x^{\frac{m}{n}})^n-1} \]

Next we can multiply the exponents \( \frac{m}{n} \) and \( n - 1 \) in the demominator

\[ \frac{d}{dx}y = \frac{m}{n} \frac{x^{m - 1}}{x^{\frac{m}{n} \cdot n - 1}} \]

This multiplication is the same as \( \frac{m}{n} \cdot \frac{n-1}{1} \) which equals \( \frac{m(n - 1)}{n} \)

\[ \frac{d}{dx}y = \frac{m}{n} \frac{x^{m - 1}}{x^{\frac{m(n - 1)}{n}}} \]

Next we can combine the two parts of the fraction by turning the demominator into a negative exponent.

\[ \frac{d}{dx}y = \frac{m}{n} x^{m - 1 - \frac{m(n - 1)}{n}} \]

To be able to do fractional division between \( \frac{m -1}{1} \) and \( \frac{m(n - 1)}{n} \) we can multiply \( \frac{m -1}{1} \) by \( n \) which results in

\[ \frac{d}{dx}y = \frac{m}{n} x^{\frac{n(m - 1)}{n} - \frac{m(n - 1)}{n}} \]

Which is the same as

\[ \frac{d}{dx}y = \frac{m}{n} x^{\frac{n(m - 1) - m(n - 1)}{n}} \]

Then multiply out to

\[ \frac{d}{dx}y = \frac{m}{n} x^{\frac{nm - n - nm + m}{n}} \]

We simplifies to

\[ \frac{d}{dx}y = \frac{m}{n} x^{\frac{m - n}{n}} \]

The fractional exponent is equivalent to

\[ \frac{d}{dx}y = \frac{m}{n} x^{\frac{m}{n} - \frac{n}{n}} \]

And \( \frac{n}{n} \) is equivalent to \( 1 \) which means this resolves to

\[ \frac{d}{dx}y = \frac{m}{n} x^{\frac{m}{n} - 1} \]

This is exactly the derivative that is show at the top of this section and is the same as the \( nx^{n-1} \) rule, thus proving the rule works even for rational exponents.