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derivative

Are you familiar with the product and chain rules?

yes...

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

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

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

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

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

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.

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

thanks!

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

is that already the answer?

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