Question

Find the derivative.
y= (cos2x)^4 - (sin2x)^4

Answer 1

\[y = [\cos (2x)]^4 - [\sin (2x)]^4\]
is that the question?

Answer 2

y'(x) = -4 sin(4 x)

Answer 3

y'(x) = 2 i e^(4 i x)-2 i e^(-4 i x)

Answer 4

see here
http://www.wolframalpha.com/input/?i=Find+the+derivative.+y%3D+%28cos2x%29%5E4+-+%28sin2x%29%5E4

Answer 5

ans)y'(x) = -8 (sin(2 x) cos^3(2 x)+sin^3(2 x) cos(2 x))

Answer 6

@lgbasallote is it correct??

Answer 7

nope

Answer 8

@lgbasallote ,yes

Answer 9

here's a hint.... i'll do the derivative of \[[\cos (2x)]^4\]
for you
\[\implies 4[\cos (2x)]^3 \times \frac{d}{dx} [\cos (2x)]\]
\[\implies 4[\cos (2x)]^3 \times [-\sin (2x)] \times \frac{d}{dx} (2x)\]
\[\implies 4[\cos (2x)]^3 \times -\sin (2x) \times 2\]
\[\implies -8 [\cos (2x)]^3 \sin (2x)\]
does that help?

Answer 10

@camille08 did u understand???????

Answer 11

then the derivative of sin2x^4 is 8 (cos 2x)(sin 2x)^3
then we are going to perform the operation.. how

Answer 12

well you can leave it as is

Answer 13

like this:
\[\large -8[\cos (2x)]^3 \sin (2x) - 8 \cos (2x) [\sin (2x)]^3\]

Answer 14

how are we going to combine that

Answer 15

like i said you can leave it like that...

Answer 16

is there an answer you're aiming for it to look like?

Answer 17

-4sin4x..

Answer 18

factor out..
\[-8 \sin (2x) \cos (2x) [ \cos^2 (2x) + \sin^2 (2x)]\]
\[\implies -8\sin (2x) \cos (2x)\]
\[\implies -4 [2\sin (2x)\cos (2x)]\]
\[\implies -4\sin (2(2x))\]
\[\implies -4\sin (4x)\]
does that help?

Answer 19

@lgbasallote , thank you..

Answer 20

welcome :)