What is the derivative of #e^(pi)+pi^(x)#?

1 Answer
Aug 4, 2015

#pi^x lnpi#

Explanation:

#e^pi# is a constant

(it's near #2.7^3# using #e~~2.7# and #pi~~3#)

Because it is a constant its derivative is #0#.

(You could use the power rule for #d/dx(u^pi)# to get #pie^(pi-1)*d/dx(e)#, but since #d/dx(e) = 0#, you still end up with #0#. Similarly with treating this as an exponential function you need to multiply by the derivative of the exponent, which is #d/dx(pi) = 0#)

#pi^x# is an exponential function with base #pi#, so its derivative uses the rule:

#d/dx(b^x) = b^x lnb#

So:

#d/dx(pi^x) = pi^x lnpi#