Question

How do you differentiate `y=(5/2)^cosx`?

Answer 1

`y' = - (frac(5)(2))^(cos(x)) ln(frac(5)(2)) sin(x)`

Explanation:

We have: `y = (frac(5)(2))^(cos(x))`

This function can be differentiated using the "chain rule".

Let `u = cos(x)` and `v = (frac(5)(2))^(u)`:

`Rightarrow y' = u' cdot v'`

`Rightarrow y' = - sin(x) cdot (frac(5)(2))^(cos(x)) cdot ln(frac(5)(2))`

`Rightarrow y' = - (frac(5)(2))^(cos(x)) ln(frac(5)(2)) sin(x)`