Let

f(x) = (7^(x))* (log base 3 of(x))
f '(x) =

you just need to memorize a couple more formulas:

you know that if
f = e^x
f' = e^x

f = a^x
f' = a^x ln(a)

logbx = ln(x)/ln(b)

f = logbx
f' = 1/ln(b) * 1/x

so, for f above,

f' = 7^x ln7 log3x + 7^x * 1/xln3
= 7^x ln7 lnx/ln3 + 7^x *1/xln3
= 7^x/xln3 (x ln7 lnx + 1)