Question

d/dx logx x^-1 anyone help?​

Answer 1

here

The answer ddx[log(x−1)]=1x−1⋅ddx[x−1]=1x−1

Explanation:

show that

ddx[loga(u)]=1u⋅lna⋅dudx

ddx[log(x−1)]=1x−1⋅ddx[x−1]=1x−1