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}

Note by Gordon Chan
3 months, 2 weeks 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.

Raghu Alluri - 2 months, 1 week ago

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Oops. Thanks for pointing out.

Gordon Chan - 2 months, 1 week 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 - 2 months, 1 week ago

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

Gordon Chan - 2 months, 1 week ago

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