Question

求下列导数

Answer 1

可以用复合函数的求导法.就是所谓的「链式法则」
y = u^v
dy/dx = d(u^v)/du * du/dx + d(u^v)/dv * dv/dx
d(u^v)/du,这里的v是常量,u是自变量；d(u^v)/dv,这里的u是常量,v是自变量
= v * u^(v - 1) * du/dx + u^v * lnu * dv/dx
= v/u * u^v * du/dx + u^v * lnu * dv/dx
= u^v * [(v/u)(du/dx) + (lnu)(dv/dx)]

1. y'=ln(cos(x))*cos(x)^x - x*cos(x)^(x - 1)*sin(x)

2. y'=ln(x^2 + 1)*(x^2 + 1)^tan(x)*(tan(x)^2 + 1) + 2*x*tan(x)*(x^2 + 1)^(tan(x) - 1)

3. y'=
   

x^x*x^(x^x - 1) + x*x^(x - 1) + x^x*ln(x) + x^(x^x)*ln(x)*(x*x^(x - 1) + x^x*ln(x)) + 1

  4.y'=[(2*x - 1)*(1 - x^2)^(1/2)]/(x^2 + 1)^(1/3) + (x*(- x^2 + x))/[(1 - x^2)^(1/2)*(x^2 + 1)^(1/3)] + [2*x*(- x^2 + x)*(1 - x^2)^(1/2)]/(3*(x^2 + 1)^(4/3))

以上结果通过matlab验证。