Question

How do you use the chain rule to differentiate `ln(-cosx)`?

Answer 1

`d/dxln(-cosx)=-tanx`

Explanation:

The Chain Rule, when applied to logarithms, tells us that

`d/dxln(u)=1/lnu*(du)/dx, u` is some function in terms of `x.` Here, `u=-cosx,` so we have

`d/dxln(-cosx)=1/(-cosx)*d/dx(-cosx)`

`d/dxln(-cosx)=sinx/(-cosx)`

`d/dxln(-cosx)=-tanx`