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

1 Answer
Mar 6, 2015

#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#.