Question

differentiate
f(x)=cos(3x)sin^2(4x)

Answer 1

is the answer
-sin(3x)sin(8x)

Answer 2

Hmmm did you forget to apply the product rule? <:o

Answer 3

yes i did

Answer 4

Is the correct answer then
(-sin(3x))(sin^2(4x))+(cos(3x))(sin(8x))

Answer 5

\[\large f(x)=\cos(3x)\sin^2(4x)\]\[\large f'(x)=(\cos(3x))'\sin^2(4x)+\cos(3x)(\sin^2(4x))'\]
\[\large = (-3\sin(3x))\sin^2(4x)+\cos(3x)(2\sin(4x)\cos(4x)(4))\]
\[\large =-3\sin(3x)\sin^2(4x)+8\cos(3x)\sin(4x)\cos(4x)\]
\[\large =-3\sin(3x)\sin^2(4x)+4\cos(3x)\sin(8x)\]

Hmm I think you forgot the 4 that comes out from differentiating the innermost function of the squared term.
Looks good besides that though! :)

Answer 6

thanks

Answer 7

Oh and you missed the 3 on the first term also, dont forget to do the chain rule multiple times if you need to :D
Multiply by (3x)' on the first term, and (4x)' on the second
Understand where those come from?

Answer 8

yes i see it now thanks i knew that i was forgetting something