Question

How do you use the chain rule to differentiate `log_13(-cosx)`?

Answer 1

`-1/ln(13)*tan(x)`

Explanation:

We Need that

`log_(a)(b)=ln(b)/ln(a)`
so we write

`log_(13)(-cos(x))=ln(-cos(x))/ln(13)` ad we get by the chain rule

`1/ln(13)*1/(-cos(x))*sin(x)`
and this is equal to

`-1/ln(13)*tan(x)`