y＝lncose的x的三次方求二阶求导

Question

如题所述

Answer 1

y=ln{cos[e^(x^3)] }
y'
= {1/cos[e^(x^3)] } .( -sin[e^(x^3)] } . [ 3x^2.e^(x^3) ]
= -3x^2. tan[e^(x^3)] . e^(x^3)
y''
=-3{2x.tan[e^(x^3)]+ x^2.(sec[e^(x^3)])^2.(3x^2)+ x^2.tan[e^(x^3)].(3x^2)}. e^(x^3)
=-3{2x.tan[e^(x^3)]+ 3x^4(sec[e^(x^3)])^2+ 3x^4.tan[e^(x^3)] }. e^(x^3)
=-3x.{2tan[e^(x^3)]+ 3x^3(sec[e^(x^3)])^2+ 3x^3.tan[e^(x^3)] }. e^(x^3)