# What is the derivative of x^n?

Oct 10, 2014

For the function $f \left(x\right) = {x}^{n}$, n should not equal 0, for reasons which will become clear. n should also be an integer or a rational number (i.e. a fraction).

The rule is:
$f \left(x\right) = {x}^{n} \implies f ' \left(x\right) = n {x}^{n - 1}$

In other words, we "borrow" the power of x and make it the coefficient of the derivative, and then subtract 1 from the power.

$f \left(x\right) = {x}^{2} \implies f ' \left(x\right) = 2 {x}^{1}$
$f \left(x\right) = {x}^{7} \implies f ' \left(x\right) = 7 {x}^{6}$
$f \left(x\right) = {x}^{\frac{1}{2}} \implies f ' \left(x\right) = \frac{1}{2} \cdot {x}^{- \frac{1}{2}}$

As I mentioned, the special case is where n=0. This means that
$f \left(x\right) = {x}^{0} = 1$
We can use our rule and technically get the right answer:
$f ' \left(x\right) = 0 {x}^{-} 1 = 0$
However, later on down the track, we will run into complications when we try to use the inverse of this rule.

Oct 9, 2015

${y}^{'} = n {x}^{n - 1}$

Below are the proofs for every numbers, but only the proof for all integers use the basic skillset of the definition of derivatives. The proof for all rationals use the chain rule and for irrationals use implicit differentiation.

#### Explanation:

That being said, I'll show them all here, so you can understand the process. Beware that it $w i l l$ be fairly long.

From $y = {x}^{n}$, if $n = 0$ we have $y = 1$ and the derivative of a constant is alsways zero.

If $n$ is any other positive integer we can throw it in the derivative formula and use the binomial theorem to solve the mess.

$y = {\lim}_{h \rightarrow 0} \frac{{\left(x + h\right)}^{n} - {x}^{n}}{h}$
$y = {\lim}_{h \rightarrow 0} \frac{{x}^{n} + {\Sigma}_{i = 1}^{n} \left({K}_{i} \cdot {x}^{n - i} {h}^{i}\right) - {x}^{n}}{h}$

Where ${K}_{i}$ is the appropriate constant

$y = {\lim}_{h \rightarrow 0} {\Sigma}_{i = 1}^{n} \frac{{K}_{i} \cdot {x}^{n - i} {h}^{i}}{h}$

Dividing that $h$

$y = {\lim}_{h \rightarrow 0} {\Sigma}_{i = 1}^{n} {K}_{i} \cdot {x}^{n - i} {h}^{i - 1}$

We can take out the first term from the sum

$y = {\lim}_{h \rightarrow 0} {K}_{1} \cdot {x}^{n - 1} + {\Sigma}_{i = 2}^{n} {K}_{i} {x}^{n - i} {h}^{i - 1}$

Taking the limit, everything else still in the sum goes to zero. Calculating ${K}_{1}$ we see that it equals $n$, so

$y = {K}_{1} \cdot {x}^{n - 1} = n {x}^{n - 1}$

For $n$ that are negative integers it's a bit more complicated. Knowing that ${x}^{-} n = \frac{1}{x} ^ b$, such that $b = - n$ and therefore is positive.

$y = {\lim}_{h \rightarrow 0} \frac{1}{h} \left(\frac{1}{x + h} ^ b - \frac{1}{x} ^ b\right)$
$y = {\lim}_{h \rightarrow 0} \frac{1}{h} \left(\frac{{x}^{b} - {\left(x + h\right)}^{b}}{{x}^{b} {\left(x + h\right)}^{b}}\right)$
$y = {\lim}_{h \rightarrow 0} \frac{1}{h} \left(\frac{{x}^{b} - {x}^{b} - {\Sigma}_{i = 1}^{b} {K}_{i} {x}^{b - i} {h}^{i}}{{x}^{b} {\left(x + h\right)}^{b}}\right)$
$y = {\lim}_{h \rightarrow 0} \left(\frac{- {\Sigma}_{i = 1}^{b} {K}_{i} {x}^{b - i} {h}^{i - 1}}{{x}^{b} {\left(x + h\right)}^{b}}\right)$

Take out the first term

$y = {\lim}_{h \rightarrow 0} \left(\frac{- {K}_{1} {x}^{b - 1} - {\Sigma}_{i = 2}^{b} {K}_{i} {x}^{b - i} {h}^{i - 1}}{{x}^{b} {\left(x + h\right)}^{b}}\right)$

Take the limit, Where ${K}_{1} = b$, subsituting that back to $n$

$y = - {K}_{1} {x}^{b - 1} / \left({x}^{b} \cdot {x}^{b}\right) = - {K}_{1} {x}^{b - 1 - 2 b} = - {K}_{1} {x}^{- b - 1} = n {x}^{n - 1}$

For rationals we need to use the chain rule. I.e.: ${\left[f \left(g \left(x\right)\right)\right]}^{'} = {f}^{'} \left(g \left(x\right)\right) {g}^{'} \left(x\right)$

So, knowing that ${x}^{\frac{1}{n}} = \sqrt[n]{x}$ and assuming $n = \frac{1}{b}$ we have

${\left({x}^{n}\right)}^{b} = x$

If $b$ is even, the answer is technically $| x |$ but this is close enough for our purposes

So, using the chain rule we have

${\left[{x}^{n}\right]}^{'} = \frac{1}{b {x}^{n b - n}} = \frac{1}{b {x}^{1 - n}} = n {x}^{n - 1}$

And last but not least, using implicit differentiation we can prove for all real numbers, including the irrationals.

$y = {x}^{n}$
$\ln \left(y\right) = n \cdot \ln \left(x\right)$

${y}^{'} / y = \frac{n}{x}$

${y}^{'} = \frac{n {x}^{n}}{x} = n {x}^{n - 1}$