利用復(fù)合函數(shù)的求導(dǎo)法則：y= f(u),u= u(x),則 f ' (x) = f '(u) * u'(x)
y = cot ( lnx),設(shè)：y = cot u,u= lnx
y ' = (cotu)' = ( sinu/ cosu) ' = [ (sinu)' *cosu - sinu * ( cosu)' ]/ (cosu)^2
= (cos^2u + sin^2u) /( cos^2u) = 1/ cos^2u
u ' = (lnx)' =1/x
所以：y ' = 1/ ([cos^2( lnx)] * x)還有一個(gè)問(wèn)題，f(x)=sinx/2xx<0=a+x ,x>o試求alimx-0 f(x)存在 急啊，謝謝了哈f(x)= sinx/ (2x)= (sinx/ x) * (1/2)那么 (sinx/ x) 的極限是1， 所以f(x)的極限是1/2 所以 a + 0 (注意x=0）= 1/2, 所以 a= 1/2
求函數(shù)y=cot(lnx)的導(dǎo)數(shù)
求函數(shù)y=cot(lnx)的導(dǎo)數(shù)
f(x)=sinx/2x xo
試求a lim
x-0 f(x)存在
f(x)=sinx/2x xo
試求a lim
x-0 f(x)存在
數(shù)學(xué)人氣：378 ℃時(shí)間：2020-05-27 20:19:00
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