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

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tkhunnyBest ResponseYou've already chosen the best response.1
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?
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
differentiate Y=cos4 t sin4 t
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
Is this cos(4t) or \[\cos^4t\]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
We're looking for \(\dfrac{dY}{dt}\)? Can you find \(\dfrac{d}{dt}\cos^{4}(t)\)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
What do you get for that?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Conversation Overlap Misunderstanding... What do you get for \(\dfrac{d}{dt}\cos^{4}(t)\)
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
its our assignment in differential calculus...
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Jay you didn't answer the question kirby asked. Is it suppose to be \(\cos(4t)\) or \(\cos^4(t)\)?
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
i dont know how to type that
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
Just think of it as the chain rule. You can move the exponent if it confuses you: \[\frac{d}{dt}\cos^4t=\frac{d}{dt}(\cos t)^4 = 4(\cos t)^3(\sin t)\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
sin t comes from the fact that it is the derivative of cos t by using the chain rule.
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
the book has an answer of 2sin 2 t
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
Just use the same logic on \[\frac{d}{dt}\sin^4t=4\sin^3t(\cos t)\]
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
cos4tsin4t= kirb can u give me the whole solution soo i can studied it... please
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
so now: \[4\cos^3t(\sin t)  4\sin^3t(\cos t) = 4\cos t*\sin t(\cos^t+\sin^2t)=4\cos t*\sin t(1) \]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
Now use the doubleangle formula
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
\[Since : \sin(2t)=2\sin t \cos t\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
\[4\cos t \sin t = 2(2\cos t \sin t) = 2\sin(2t)\]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
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.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
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).
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.3
The method by tkhunny is also excellent :) It is shorter but I usually just do it "straightforward" unless I'm stuck and tkhunny's method is a good trick to make the derivative a lot easier
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
I'm usually the brute force guy. Once in a while I see one!
 one year ago

jayshaneBest ResponseYou've already chosen the best response.0
how about differentiate: y=sec²xtan²x
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
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.
 one year ago
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