Power Rule

(xn)=limh0(x+h)n(xn)h=limh0(k=0n(nk)xnkhk)xnh=limh0xn+(k=1n(nk)xnkhk)xnh=limh0k=1nh(nk)xnkhk1h=limh0h(k=1n(nk)xnkhk1)h=limh0k=1n(nk)xnkhk1=limh0((n1)xn1h0+(n2)xn2h1+(n3)xn3h2+)=limh0(nxn1+h((n2)xn2+(n3)xn3h1+))=nxn1+0((n2)xn2+(n3)xn301+)=nxn1\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}

#Calculus

Note by Gandoff Tan
1 year, 11 months ago

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Comments

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.

Raghu Alluri - 1 year, 10 months ago

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Oops. Thanks for pointing out.

Gandoff Tan - 1 year, 10 months ago

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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.

Raghu Alluri - 1 year, 10 months ago

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

Gandoff Tan - 1 year, 10 months ago

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

A Former Brilliant Member - 1 year, 1 month ago

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

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

Brilliant Mathematics Staff - 1 year ago
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