求导数y=cos[In(1+3x)]

问题描述:

求导数y=cos[In(1+3x)]

y=cos[In(1+3x)]
y'=-sin[In(1+3x)][In(1+3x)]'
y'=-sin[In(1+3x)][1/(1+3x)](1+3x)'
y'=-3sin[In(1+3x)]/(1+3x)