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DERIVATIVES OF TRIGO IDENTITIES no.1)Y=cos4 t- sin4 t=-2sin 2t INEED SOLUTION PLS....T_T NID BADLY!!!!

Mathematics
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It makes no sense as it it presented. Trigonometric Identities are statements and the term "derivative" has no meaning for such a statement. Can you provide the exact wording of the problem?
differentiate Y=cos4 t -sin4 t
Is this cos(4t) or \[\cos^4t\]

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Other answers:

We're looking for \(\dfrac{dY}{dt}\)? Can you find \(\dfrac{d}{dt}\cos^{4}(t)\)
yes
What do you get for that?
ist assingment dude
Conversation Overlap Misunderstanding... What do you get for \(\dfrac{d}{dt}\cos^{4}(t)\)
its our assignment in differential calculus...
Jay you didn't answer the question kirby asked. Is it suppose to be \(\cos(4t)\) or \(\cos^4(t)\)?
the second one
i dont know how to type that
Just think of it as the chain rule. You can move the exponent if it confuses you: \[\frac{d}{dt}\cos^4t=\frac{d}{dt}(\cos t)^4 = 4(\cos t)^3(-\sin t)\]
-sin t comes from the fact that it is the derivative of cos t by using the chain rule.
the book has an answer of -2sin 2 t
Just use the same logic on \[\frac{d}{dt}\sin^4t=4\sin^3t(\cos t)\]
cos4t-sin4t= kirb can u give me the whole solution soo i can studied it... please
so now: \[-4\cos^3t(\sin t) - 4\sin^3t(\cos t) = -4\cos t*\sin t(\cos^t+\sin^2t)=-4\cos t*\sin t(1) \]
Now use the double-angle formula
\[Since : \sin(2t)=2\sin t \cos t\]
\[-4\cos t \sin t = -2(2\cos t \sin t) = -2\sin(2t)\]
THANK YOU KIRB!!!
Since we're just doing your homework, I'd do it this way. \(\cos^{4}(t) - \sin^{4}(t) = [\cos^{2}(t) - \sin^{2}(t)][\cos^{2}(t) + \sin^{2}(t)] = \cos(2x)\) It's a lot easier after that.
Sorry, not sure why I wrote 2x on the end, there. Should be 2t. If you are going to make me do ALL the work, you'll have to show me how that last step happened. There's a whole lot of stuff in there that magically turned into cos(2t).
The method by tkhunny is also excellent :) It is shorter but I usually just do it "straight-forward" unless I'm stuck and tkhunny's method is a good trick to make the derivative a lot easier
I'm usually the brute force guy. Once in a while I see one!
Hehe good one ;)
how about differentiate: y=sec²x-tan²x
Go right ahead. Let's see your first attempt. Hint: It is amazingly trivial! Remember your trigonometry. This is not much of a calculus problem.

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