derive (-3x)^(2x) cos(3x) - Sin(2x)

Question

anonymous · Answer

what do you mean derive

anonymous · Answer

i ment the derivative of that equation :)

anonymous · Answer

is it supposed to be = to sin(2x), instead of -?

anonymous · Answer

you have to derive something from something else

anonymous · Answer

no that's really -

anonymous · Answer

erm i don't know. the question doesn't make any sense

anonymous · Answer

it's like saying derive x + y. you have to derive it from something else

anonymous · Answer

you see, i'm trying to solve its roots using the newton-rhapson method and i have to get the derivative of $$f(x)=-3x ^{2x}\cos(3x) - \sin(2x)$$

anonymous · Answer

:( i don't know that method sorry

anonymous · Answer

that's ok... thanks

jack1 · Answer

$$\large if~y = (-3x)^{2x} \cos(3x) - \sin (2x)$$
$$\large then~y' = 9^x (-x)^{2 x} (2 (log(-3 x)+1) cos(3 x)-3 sin(3 x))-2 cos(2 x)$$
use chain rule and product rule.

jack1 · Answer

assume you already know base rule: derivative sin = cos, derivative cos = -sin? yeah?

jack1 · Answer

*base rules

anonymous · Answer

thanks jack1

anonymous · Answer

jack1? is this answer simplified already?

jack1 · Answer

yeah

anonymous · Answer

how do you input sec in the sci cal?