How do you differentiate #f(x)=23^x#?
1 Answer
Mar 8, 2017
#d/(dx) 23^x = 23^x ln 23#
Explanation:
Use:
#23 = e^ln(23)#
#(a^b)^c = a^(bc)" "# when#a > 0#
#d/(dx) e^x = e^x#
#d/(dx) u(v(x)) = u'(v(x))*v'(x)#
So:
#d/(dx) 23^x = d/(dx) (e^(ln 23))^x = d/(dx) e^((ln 23)x) = e^((ln 23) x) ln 23 = 23^x ln 23#
