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anonymous
 5 years ago
y=sin^4xcos^4y
anonymous
 5 years ago
y=sin^4xcos^4y

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you familiar with the product and chain rules?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, we'll use both of those to solve this. Let's start from the beginning.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This function is two functions being multiplied. (sin x)^4 and (cos x)^4 Can you tell me the derivative of each of these individually?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of (sin x)^4 is 4(sin x)3 and tfor the (cos y)^4 is 4(cos y)3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Remeber to use the chain rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x) = \sin ^{4}x\] \[f'(x) = 4\sin ^{3}x cosx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you differentiate the (cos x)^4?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The cosine term doesn't have an x, although y is a function of x, so we need implicit differentiation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[g(x) = \cos ^{4}y(x)\] \[g'(x) = 4y'\cos ^{3}y \sin y\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you put all of this together with the product rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sin^{4}x(4y ^{1}\cos ^{3}\sin y)+\cos ^{4}y(4\sin ^{3}xcosx)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The mark after the 4y is a "'", meaing the derivative of y (y prime). But there you have it. The derivative of y, with y being a function of x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please check if this is correct...because my teacher posted this as an answer..\[y ^{1}=4\sin ^{3}xcos ^{3}y[ sinx(siny)+cosy cosx]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm, I'll be back. I'll play with this and see if I can reproduce what your teacher got.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so that is what we have, but factored a bit, and without the y' term.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would ask your teacher why he didn't treat y as a function of x when taking the derivative of the cosine term. Other than that, we have exactly what he has.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please help me with this one sir...\[y=10^{\cos2x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm, it looks like you'll have to take the log of both sides, then do a derivative. I'm helping someone else in another thread, but I'll post back as soon as I find the answer to this one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, the answer to your last question is a little indepth. Do you know the derivative of a base 10 logarithm?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im sorry, i know nothing about this one...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah...It's not something most people learn. Interesting that you need to do this one. Anyway, take the log of both sides: \[\log y = \cos 2x\] Differentiate both sides: \[dy/yln10 = 2\sin2x dx\] Solve for dy/dx: \[dy/dx = 2y \ln10 \sin2x\] Substitute in for y (given initially) and simplify. Honestly, I'd check to make sure you have the problem written correctly. This isn't one that I see too often.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that already the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Close. You need to sub in for y, and then simplify a bit.
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