Question

What is the derivative of `1/(x+1)`?

Answer 1

This function can be rewritten as `(x+1)^-1`, following the law of exponentials that states `a^-n=1/a^n`.

Now, we can differentiate it using the chain rule, which, in turn, states that

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

In this case, if we rename `u=x+1`, then, we have that the original function `y=(x+1)^-1` becomes `y=u^-1` and, now, we can proceed to the chain rule steps:

`(dy)/(du)=-1*u^-2`

`(du)/(dx)=1`

Thus,

`(dy)/(dx)=(-1u^-2)(1)=-1/u^2=-1/(x+1)^2`