设f(x)=e^(-x) *ln(2-x) + (1+3x^2)^(1/2),求f'(x)

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设f(x)=e^(-x) *ln(2-x) + (1+3x^2)^(1/2),求f'(x)
设f(x)=e^(-x) *ln(2-x) + (1+3x^2)^(1/2),求f'(x)

设f(x)=e^(-x) *ln(2-x) + (1+3x^2)^(1/2),求f'(x)
复合函数求导:
f'(x)=[e^(-x) *ln(2-x)]'+ [(1+3x^2)^(1/2)]'
=[e^(-x)]'*ln(2-x)+e^(-x)*[ln(2-x)]'+[(1+3x^2)^(1/2)]'
=-e^(-x)*ln(2-x)-e^(-x)*(1/(2-x))+(1/2)(1+3x^2)^(-1/2)*6x
原理是
设f(x)=g[p(x)]
则f'(x)=g'[p(x)]*p'(x)

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)
复合函数求导:由外到内
如:
f(x)=g(q(x))
f'(x)=g'(q(x))q'(x)
===========================
f(x)=e^(-x) *ln(2-x) + (1+3x^2)^(1/2)
f'(x)=[e^(-x)]'ln(2-x)...

全部展开

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)
复合函数求导:由外到内
如:
f(x)=g(q(x))
f'(x)=g'(q(x))q'(x)
===========================
f(x)=e^(-x) *ln(2-x) + (1+3x^2)^(1/2)
f'(x)=[e^(-x)]'ln(2-x)+e^(-x)[ln(2-x)]'+[(1+3x^2)^(1/2)]'
=-e^(-x)ln(2-x)+e^(-x)(-1)/(2-x)+1/2*(1+3x^2)^(-1/2)*(1+3x^2)'
=-e^(-x)ln(2-x)-e^(-x)/(2-x)+3x/[根号(1+3x^2)]
前面的好像错了

收起

f'(x)=[e^(-x)]'* ln(2-x)+e^(-x)*[ln(2-x)]'+[(1+3x^2)]'*1/2(1+3x^2)^(1/2)
=-e^(-x) * ln(2-x) -e^(-x)/(2-x) + 3x/(1+3x^2)^(1/2)