Question

derivative of y=log(x^x)^x​

Answer 1

Answer:

good luck and keep the same as the same as the same as the same as the same as the same as the same as the same as the same as the same as the same as the same as the same as the same as

Answer 2

Answer:

x(1+logx)+logx^x

step by step explanation:

let y=log(x^x)^x

we know that log a^m=mlog a

• y=x log(x^x)
• dy/dx=x d/dx(logx^x) + logx^x d/dx(x)

d/dx(uv)=u dv/dx+v du/dx

• dy/dx=x•1/x^x •d/dx(x^x) + logx^x (1)

d/dx(logx)=1/x

• dy/dx=x•1/x^x •x^x(1+logx) + logx^x

d/dx(x^x)=x^x(1+logx)

then x^x will be cancelled

• dy/dx=x(1+logx) + logx^x