anonymous
  • anonymous
y=sin^4xcos^4y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
derivative
anonymous
  • anonymous
Are you familiar with the product and chain rules?
anonymous
  • anonymous
yes...

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

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