Question

y=sin^4xcos^4y

Answer 1

derivative

Answer 2

Are you familiar with the product and chain rules?

Answer 3

yes...

Answer 4

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

Answer 5

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?

Answer 6

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

Answer 7

Close

Answer 8

Remeber to use the chain rule.

Answer 9

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

Answer 10

Can you differentiate the (cos x)^4?

Answer 11

f′(x)=4cosy3x-siny

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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.

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

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

Answer 22

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.

Answer 23

thanks!

Answer 24

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

Answer 25

im sorry, i know nothing about this one...

Answer 26

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.

Answer 27

is that already the answer?

Answer 28

Close. You need to sub in for y, and then simplify a bit.