What is the derivative of #4^x#?

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Answer 1 · 2015-08-24T16:13:30

#d/dx(4^x) = 4^x ln4#

In general for #b>0#, we have
#d/dx(b^x) = b^x lnb#

And when we need the chain rule, we have

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

Proof of General rule

#b^x = (e^lnb)^x# #" "# (remember that #e^lnu = u#)

So
#b^x = e^(xlnb)# #" "# (because #(a^r)^s = a^(rs) = a^(sr)#)

#d/dx(e^(xlnb))# can be found by the chain rule:

#d/dx(e^(xlnb)) = e^(xlnb) d/dx(xlnb)#

# = e^(xlnb) lnb# #" "# (#lnb# is a constant)

# = b^x lnb# #" "# (reverse the first two steps.) #square#