Question

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

Answer 1

what do you mean derive

Answer 2

i ment the derivative of that equation :)

Answer 3

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

Answer 4

you have to derive something from something else

Answer 5

no that's really -

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

:( i don't know that method sorry

Answer 10

that's ok... thanks

Answer 11

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

Answer 12

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

Answer 13

*base rules

Answer 14

thanks jack1

Answer 15

jack1? is this answer simplified already?

Answer 16

yeah

Answer 17

how do you input sec in the sci cal?