Maths wizards, remind me is there a derivative of 2^3x.dx ?
Y=2^(3x). Ln both sides. Lny=ln 2^(3x). Use prop of logs. Lny=3x (ln2). Diff each side. 1/y dy/dx= 3(ln2). Solve for dy/dx. dy/dx= 3(ln2)(y). Sub orig equation in. dy/dx= 3(ln2)(2^3x)
Uhhhhh... Guess I'm not a math wizard. I'll stick with my history...
y=2^3XY ' = 2^3X ln2= 2^3X x 0.6let me know if you don't get itI'm a vampire by the way ;)
f(x)=2^3xf ' (x)= 3(2^3)f ' (x)= 3(8)f ' (x)= 24