用对数求导法求下列函数的导数？

Question

求解

Answer 1

(1) lny = ln2 + ln|x| + (1/2)ln|(1-2x)| - (1/2)ln|(1+2x)|
y'/y = 0 + 1/x +(1/2)(-2)/(1-2x) - (1/2)2/(1+2x)
= 1/x - 1/(1-2x) - 1/(1+2x)
y' = y[1/x - 1/(1-2x) - 1/(1+2x)]
= 2x√[(1-2x(/(1+2x)][1/x - 1/(1-2x) - 1/(1+2x)]
(2) 定义域 x > 0, sinx > 0.
记 u = 2x^(2x), 则 lnu = ln2 + 2xlnx,
u'/u = 0+2lnx+2 = 2(1+lnx)
u' = 2u(1+lnx) = 4x^(2x)(1+lnx)
记 v = 2(sinx)^(2x), 则 lnv = ln2 + 2xln(sinx)
v'/v = 0 + 2ln(sinx) + 2xcosx/sinx = 2ln(sinx)+2xcotx
v' = v[2ln(sinx)+2xcotx] = 4(sinx)^(2x)[ln(sinx)+xcotx]
y' = u'+v' = 4x^(2x)(1+lnx) + 4(sinx)^(2x)[ln(sinx)+xcotx]