Find the f''' of function f(x) = e2x6 + [lnx]2.
Does we need to do by part? Can someone plz give me the final answer. I do this without using the by part formula..can it be correct?
assuming you meant
f(x) = e^(2x^6) + (lnx)^2
f'(x) = 12x^5 e^(2x^6) + 2lnx/x
f"(x) = 60x^4 e^(2x^6) + (12x^5)^2 e^(2x^6) + 2/x^2 - 2lnx/x^2
= 12x^4 (12x^6 + 5) e^(2x^6) - 2lnx/x^2
f"'(x) = (48x^3 (12x^6+5) + 12x^4 (72x^5)) e^(2x^6) - 4/x^3 - 2/x^3 + 4lnx/x^3
= 48x^3 (36x^12 + 45x^6 + 5) - 6/x^3 + 4lnx/x^3