Question

高数求导数

第二张图清晰一些

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Answer 1

f(x)
=3x-lnx ; 0<x<1
=3x+lnx ; x≥1
f(1+)=f(1) = lim(x->1) (3x+lnx) = 3
f(1-)= lim(x->1-) (3x-lnx) = 3
x=1, f(x) 连续
f'(1+)
=lim(h->0+) [ 3(1+h)+ln(1+h) - f(1) ] /h
=lim(h->0+) [ 3(1+h)+ln(1+h) - 3 ] /h
=lim(h->0+) [ 3h+ln(1+h) ] /h
=lim(h->0+) ( 3h+h ) /h
=4
f'(1-)
=lim(h->0-) [ 3(1+h)-ln(1+h) - f(1) ] /h
=lim(h->0-) [ 3(1+h)-ln(1+h)-3 ] /h
=lim(h->0-) ( 3h-h ) /h
=2
=> f'(1) 不存在
ie
f'(x)
=3- 1/x ; 0<x<1
=3+1/x ; x>1
f'(1) 不存在

Answer 2

详细过程如图所示，希望能帮到你