Trying to differentiate using function of a function/Chain rule

Mike2793
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Homework Statement



Using the function of a function formula differentiate the following:

y = (cos 2x)3

Homework Equations



dy/dx = dy/du * du/dx

The Attempt at a Solution



y = (cos 2x)3

u = cos 2x
y = u3

dy/du = -2sin2x
du/dy = 3u2

dy/dx = dy/du * du/dx

dy/dx = -2sin2x * 3(cos2 2x)

= -6sin2x cos22x

Hi, I've been racking my brain over this question for sometime now, I've attempted it multiple times getting it wrong over and over again. I've got this answer, but, I'm really not confident that it is right. Can you guys take a look at it and work some magic? It's function of a function, which I think is more commonly known as the Chain rule.

Thank you.
 
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That looks right to me.
 
Seriously?

I'm speechless!
 
Mike2793 said:
dy/du = -2sin2x
du/dy = 3u2
should read as:
\frac{dy}{du} = 3u^2 and
\frac{du}{dx} = -2\sin 2x
but other than that, it looks fine.
 
Haha, there's my attention to detail. Thanks very much, guys, I can finally lay this to rest.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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