# Power Rule

\begin{aligned} \big({ x }^{ n }\big)\prime & = \lim _{ h\rightarrow 0 }{ \frac { { (x+h) }^{ n }-\big({ x }^{ n }\big) }{ h } } \\ \quad & = \lim _{ h\rightarrow 0 }{ \frac { \bigg(\displaystyle \sum _{ k=0 }^{ n }{ \binom{n}{k} { x }^{ n-k }{ h }^{ k } } \bigg)-{ x }^{ n } }{ h } } \\ \quad & = \lim _{ h\rightarrow 0 }{ \frac { { x }^{ n }+\bigg(\displaystyle \sum _{ k=1 }^{ n }{ \binom{n}{k} { x }^{ n-k }{ h }^{ k } } \bigg)-{ x }^{ n } }{ h } } \\ \quad & = \lim _{ h\rightarrow 0 }{ \frac {\displaystyle \sum _{ k=1 }^{ n }{ h\binom{n}{k} { x }^{ n-k }{ h }^{ k-1 } } }{ h } } \\ \quad & = \lim _{ h\rightarrow 0 }{ \frac { h\bigg(\displaystyle \sum _{ k=1 }^{ n }{ \binom{n}{k} { x }^{ n-k }{ h }^{ k-1 } } \bigg) }{ h } } \\ \quad & = \lim _{ h\rightarrow 0 }{ \sum _{ k=1 }^{ n }{ \binom{n}{k} { x }^{ n-k }{ h }^{ k-1 } } } \\ \quad & = \lim _{ h\rightarrow 0 }{ \bigg(\binom{n}{1} { x }^{ n-1 }{ h }^{ 0 }+\binom{n}{2} { x }^{ n-2 }{ h }^{ 1 }+\binom{n}{3} { x }^{ n-3 }{ h }^{ 2 }+\cdots \bigg) } \\ \quad & = \lim _{ h\rightarrow 0 }{ \Bigg(n{ x }^{ n-1 }+h\bigg(\binom{n}{2} { x }^{ n-2 }+\binom{n}{3} { x }^{ n-3 }{ h }^{ 1 }+\cdots \bigg)\Bigg) } \\ \quad & = n{ x }^{ n-1 }+0\bigg(\binom{n}{2} { x }^{ n-2 }+\binom{n}{3} { x }^{ n-3 }{ 0 }^{ 1 }+\cdots \bigg) \\ \quad & = \boxed { n{ x }^{ n-1 } } \end{aligned}

Note by Gandoff Tan
2 years, 1 month ago

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I see an error in the fifth line of your reasoning to the power rule. Since you are summing the h's over and over again you can simply factor it out so once it is factored out the h no longer appears in the summand.

- 2 years ago

Oops. Thanks for pointing out.

- 2 years ago

No Problem. I have one more question, do you like the Joy Of Problem solving course. I seem to love it. I want to know from you if there are other problem solving courses too on this. Thanks.

- 2 years ago

I like it. I think you can try the 'Ace the AMC' course.

- 2 years ago

I wonder if there is any way to show this without calculus? @Gandoff Tan

- 1 year, 2 months ago

@Brilliant Mathematics, @Creative Biogene is not talking about mathematics.

- 1 year, 2 months ago

Thank you again! We will take action on this account.

Staff - 1 year, 2 months ago