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Why is this true?

Differentiation of xn

 

f(x) = xn

f'(x) = nx(n-1)

 

Can someone explain? Thanks.

 Jul 25, 2017
 #1
avatar+9675 
+3

Use the first principles.

ddxxn=limh0(x+h)nxnh=limh0xn+(n1)xn1h+...xnhother terms are negligible as they evaluate to 0 after taking limit.=limh0nhxn1h+lots of negligible terms which evaluates to 0=limh0nxn1=nxn1 YAY

 Jul 25, 2017
 #2
avatar+26396 
+4

Simple rules of differentiation

Why is this true?

Differentiation of xn

f(x) = xn

f'(x) = nx(n-1)

 

difference quotient:

ΔyΔx=(x+h)nxnhΔyΔx=(n0)xn+(n1)xn1h+(n2)xn2h2+(nn)hnxnh|(n0)=1ΔyΔx=xn+(n1)xn1h+(n2)xn2h2+(nn)hnxnhΔyΔx=(n1)xn1h+(n2)xn2h2+(nn)hnhΔyΔx=(n1)xn1+(n2)xn2h1+(nn)hn1|(n1)=nΔyΔx=nxn1+(n2)xn2h+(nn)hn1

 

 

differential quotient:

δyδx=f(x)=limh0(nxn1+(n2)xn2h+(nn)hn1)f(x)=nxn1+(n2)xn2×0+(nn)×0n1f(x)=nxn1+0+0f(x)=nxn1

 

 

laugh

 Jul 25, 2017
 #3
avatar+9675 
0

Your answer is nice :O

Seems like I still need to learn how to show my points... :D

MaxWong  Jul 25, 2017
 #4
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0

Love you two so much!

 Jul 25, 2017

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