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

1. anonymous

derivative

2. anonymous

Are you familiar with the product and chain rules?

3. anonymous

yes...

4. anonymous

Alright, we'll use both of those to solve this. Let's start from the beginning.

5. 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?

6. anonymous

the derivative of (sin x)^4 is 4(sin x)3 and tfor the (cos y)^4 is 4(cos y)3

7. anonymous

Close

8. anonymous

Remeber to use the chain rule.

9. anonymous

$f(x) = \sin ^{4}x$ $f'(x) = 4\sin ^{3}x cosx$

10. anonymous

Can you differentiate the (cos x)^4?

11. anonymous

f′(x)=4cosy3x-siny

12. anonymous

The cosine term doesn't have an x, although y is a function of x, so we need implicit differentiation.

13. anonymous

$g(x) = \cos ^{4}y(x)$ $g'(x) = -4y'\cos ^{3}y \sin y$

14. anonymous

Can you put all of this together with the product rule?

15. anonymous

$\sin^{4}x(-4y ^{1}\cos ^{3}\sin y)+\cos ^{4}y(4\sin ^{3}xcosx)$

16. 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.

17. anonymous

please check if this is correct...because my teacher posted this as an answer..$y ^{1}=4\sin ^{3}xcos ^{3}y[ sinx(-siny)+cosy cosx]$

18. anonymous

Hmm, I'll be back. I'll play with this and see if I can reproduce what your teacher got.

19. anonymous

Ok, so that is what we have, but factored a bit, and without the y' term.

20. 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.

21. anonymous

please help me with this one sir...$y=10^{\cos2x}$

22. 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.

23. anonymous

thanks!

24. anonymous

So, the answer to your last question is a little indepth. Do you know the derivative of a base 10 logarithm?

25. anonymous

26. 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.

27. anonymous