if f(x)= from n=0 to ∞ Σcn (x-a)
prove f ' (x)= from n=0 to ∞ Σ ncn (x-a)^n-1
please steps by steps !
題目應該是這樣吧?
if f(x)= from n=0 to ∞ Σcn (x-a)^n
prove f'(x)= from n=0 to ∞ Σ ncn (x-a)^(n-1)
f(x)=c1(x-a)+c2(x-a)^2+c3(x-a)^3+...
根據(d/dx)[f(x)+g(x)]=f'(x)+g'(x)
(d/dx)(x-a)=1, f'(x)=c1(d/dx)(x-a)+c2(d/dx)(x-a)^2+c3(d/dx)(x-a)^3+...
再根據(d/dx)f(g(x))=f'(g(x))*g'(x) (chain rule)
(d/dx)(x-a)^n=n(x-a)^(n-1)*1=n(x-a)^(n-1)
綜合以上
f'(x)
=(d/dx)[from n=0 to ∞ Σ cn(x-a)^n]
=from n=0 to ∞ Σ cn(d/dx)(x-a)^n
=from n=0 to ∞ Σ ncn(x-a)^(n-1)
希望以上回答有幫助到您!
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
題目應該是這樣吧?
if f(x)= from n=0 to ∞ Σcn (x-a)^n
prove f'(x)= from n=0 to ∞ Σ ncn (x-a)^(n-1)
f(x)=c1(x-a)+c2(x-a)^2+c3(x-a)^3+...
根據(d/dx)[f(x)+g(x)]=f'(x)+g'(x)
(d/dx)(x-a)=1, f'(x)=c1(d/dx)(x-a)+c2(d/dx)(x-a)^2+c3(d/dx)(x-a)^3+...
再根據(d/dx)f(g(x))=f'(g(x))*g'(x) (chain rule)
(d/dx)(x-a)^n=n(x-a)^(n-1)*1=n(x-a)^(n-1)
綜合以上
f'(x)
=(d/dx)[from n=0 to ∞ Σ cn(x-a)^n]
=from n=0 to ∞ Σ cn(d/dx)(x-a)^n
=from n=0 to ∞ Σ ncn(x-a)^(n-1)
希望以上回答有幫助到您!