Question

What is the derivative of `2^(cosx)`?

Answer 1

`(dy)/(dx)=-log2sinx2^(cosx)`

Explanation:

Making `y=2^(cosx)`and applying `log` to both sides

`log y = cosx log2` Deriving now

`(dy)/y=-log2sinx dx` Rearranging

`(dy)/(dx)=-log2ysinx=-log2sinx2^(cosx)`