How do you derive this equation:

y = sin ^ 3 (cosx^3)

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- Apr 5th 2009, 06:49 PMmathamatics112Trig differentation
How do you derive this equation:

y = sin ^ 3 (cosx^3) - Apr 5th 2009, 07:11 PMmatheagle
This is the chain rule if the cosine is inside the sine, which what this seems to be.

- Apr 5th 2009, 07:16 PMReckoner
You need to use the chain rule several times:

$\displaystyle y=\sin^3\cos x^3=\left(\sin\cos x^3\right)^3$

$\displaystyle \Rightarrow\frac{dy}{dx}=3\sin^2\cos x^3\cdot\frac d{dx}\left[\sin\cos x^3\right]$

$\displaystyle =3\sin^2\cos x^3\left(\cos\cos x^3\cdot\frac d{dx}\left[\cos x^3\right]\right)$

And so on. Can you continue? - Apr 5th 2009, 07:17 PMmathamatics112
y= sin^3( cos x^3)

= 3 sin ^ 2 ( cos x^3)(coz(cosx^3))(-sinx^3)(3x^2)

= 9 x^2(sin^2)(cos(cos x^3))(-sin x^3))

are my steps right? - Apr 5th 2009, 07:23 PMReckoner
- Apr 5th 2009, 07:27 PMmathamatics112
i'm not seeing it :(

- Apr 5th 2009, 07:33 PMReckoner