# Power Rule

## Key Questions

• We can use the Power Rule and the Difference Quotient ( First Principles ).

Power Rule

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

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

Difference Quotient ( First Principles )

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$f \left(x\right) = \sqrt{x}$

$f \left(x + h\right) = \sqrt{x + h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{x + h - x}{h \cdot \left(\sqrt{x + h} + \sqrt{x}\right)}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{h}{h \cdot \left(\sqrt{x + h} + \sqrt{x}\right)}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}$

$f ' \left(x\right) = \frac{1}{\sqrt{x + 0} + \sqrt{x}}$

$f ' \left(x\right) = \frac{1}{\sqrt{x} + \sqrt{x}}$

$f ' \left(x\right) = \frac{1}{2 \sqrt{x}}$

Please see the videos below and a similar question.

• You could use the quotient rule or you could just manipulate the function to show its negative exponent so that you could then use the power rule.

I will convert the function to its negative exponent you make use of the power rule.

$y = \frac{1}{\sqrt{x}} = {x}^{- \frac{1}{2}}$

Now bring down the exponent as a factor and multiply it by the current coefficient, which is 1, and decrement the current power by 1.

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

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

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.

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

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

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