泛白 asked in 科學數學 · 9 years ago

微積分chain rule應用

f(x)=∫ (從x到x^2) e^(t^2) dt,求f'(0)=?

3 Answers

Rating
  • 9 years ago
    Favorite Answer

    圖片參考:http://imgcld.yimg.com/8/n/AF03209975/o/1611072403...

    微積分基本定理 + chain rule

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  • 9 years ago

    令 G(t)=∫ e^(t^2) dt

    G '(t)=e^(t^2)

    by chain rule

    G'(x)=e^(t^2)*(dt/dx)

    so

    f(x)=∫ (從x到x^2) e^(t^2) dt

    f '(x)=e^(x^2)^2*(x^2)'-e^(x^2)*(x)'

    =(2x)*e^(x^4)-e^(x^2)

    f '(0)=(2*0)*e^(0^4)-e^(0^2)

    =0-1=-1

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  • 興豪
    Lv 5
    9 years ago

    Set F(x)=∫[Ψ(x),φ(x)] f(t) dt ; F'(x) = f(φ(x))φ’(x) - f(Ψ(x))Ψ’(x)

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