Question

D/dx log (x-1) ?

Answer 1

The answer `d/dx[log (x-1)]=1/(x-1)*d/dx[x-1]=1/(x-1)`

Explanation:

show that

`d/dx[log_a(u)]=1/[u*lna]*(du)/dx`

`d/dx[log (x-1)]=1/(x-1)*d/dx[x-1]=1/(x-1)`

Answer 2

`1/(x-1)`

Explanation:

Given: `d/dx(log(x-1))`.

Use the chain rule, which states that,

`dy/dx=dy/(du)*(du)/dx`

Let `u=x-1,:.(du)/dx=1`.

Then `y=logu,dy/(du)=1/u`.

Combining, we get:

`dy/dx=1/u*1`

`=1/u`

Substitute back `u=x-1` to get the final answer:

`=1/(x-1)`