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  • f(x)=ln(x+√1+x^2) 求導

    f(x)=ln(x+√1+x^2) 求導
    數(shù)學人氣:384 ℃時間:2020-02-20 16:34:07
    優(yōu)質(zhì)解答
    f(x)=ln(x+√1+x^2)
    f'(x)=1/(x+√(1+x^2) *(x+√1+x^2)'
    =1/(x+√(1+x^2)*(1+(√1+x^2)'
    =1/(x+√(1+x^2)*(1+1/2*√(x^2+1) *(x^2)')
    =1/(x+√(1+x^2)*(1+1/2*√(x^2+1) *2x)
    =(1+x/√(x^2+1))/(x+√(1+x^2) 分子分母乘x-√(1+x^2)
    =(1+x/√(x^2+1)(x-√(x^2+1)/(x^2-1-x^2)
    =(1+x/√(x^2+1)(√(x^2+1)-x)
    =√(x^2+1)-x+x-x^2/√(x^2+1)
    =√(x^2+1-x^2√(x^2+1)/(x^2+1)
    =√(x^2+1)(1-x^2/(x^2+1))
    =√(x^2+1)(x^2+1-x^2)/(x^2+1)
    =√(x^2+1)/(x^2+1)
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