y=sin^4xcos^4y

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y=sin^4xcos^4y

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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derivative
Are you familiar with the product and chain rules?
yes...

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Alright, we'll use both of those to solve this. Let's start from the beginning.
This function is two functions being multiplied. (sin x)^4 and (cos x)^4 Can you tell me the derivative of each of these individually?
the derivative of (sin x)^4 is 4(sin x)3 and tfor the (cos y)^4 is 4(cos y)3
Close
Remeber to use the chain rule.
\[f(x) = \sin ^{4}x\] \[f'(x) = 4\sin ^{3}x cosx\]
Can you differentiate the (cos x)^4?
f′(x)=4cosy3x-siny
The cosine term doesn't have an x, although y is a function of x, so we need implicit differentiation.
\[g(x) = \cos ^{4}y(x)\] \[g'(x) = -4y'\cos ^{3}y \sin y\]
Can you put all of this together with the product rule?
\[\sin^{4}x(-4y ^{1}\cos ^{3}\sin y)+\cos ^{4}y(4\sin ^{3}xcosx)\]
The 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.
please check if this is correct...because my teacher posted this as an answer..\[y ^{1}=4\sin ^{3}xcos ^{3}y[ sinx(-siny)+cosy cosx]\]
Hmm, I'll be back. I'll play with this and see if I can reproduce what your teacher got.
Ok, so that is what we have, but factored a bit, and without the y' term.
I 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.
please help me with this one sir...\[y=10^{\cos2x}\]
Hmm, 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.
thanks!
So, the answer to your last question is a little indepth. Do you know the derivative of a base 10 logarithm?
im sorry, i know nothing about this one...
Yeah...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.
is that already the answer?
Close. You need to sub in for y, and then simplify a bit.

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