Question

How do you find the derivative of `f(x)=pi^cosx`?

Answer 1

`f'(x)=-(sinx )pi^cosx ln pi`
(I use `ln` for natural logarithm.)

To get this answer, use the formula for differentiating exponential functions with base `a` and use the chain rule.

The derivative of `a^x` is `a^xlna`
(or `a^xloga` if you use `log` for the natural logarithm).

So, for example, the derivative of `g(x)=pi^x` would be `g'(x)=pi^x ln pi`

You function `f(x)=pi^cosx` doesn't have exponent simply `x`.
It is of the form `y=pi^u` the derivative of which is `pi^u lnpi ((du)/dx)`.
We need the chain rule here, because `u!=x`.

Because your function has `u=cosx`, you'll use `(du)/dx=-sinx`.

`f'(x)=pi^cosx(ln pi)(-sinx)=-(sinx )pi^cosx ln pi`.