# Trig differentation

• Apr 5th 2009, 06:49 PM
mathamatics112
Trig differentation
How do you derive this equation:

y = sin ^ 3 (cosx^3)
• Apr 5th 2009, 07:11 PM
matheagle
This is the chain rule if the cosine is inside the sine, which what this seems to be.
• Apr 5th 2009, 07:16 PM
Reckoner
Quote:

Originally Posted by mathamatics112
How do you derive this equation:

y = sin ^ 3 (cosx^3)

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 PM
mathamatics112
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 PM
Reckoner
Quote:

Originally Posted by mathamatics112
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))

Aren't you missing something in that last line?
• Apr 5th 2009, 07:27 PM
mathamatics112
i'm not seeing it :(
• Apr 5th 2009, 07:33 PM
Reckoner
Quote:

Originally Posted by mathamatics112
i'm not seeing it :(

Did you look at the part that I highlighted? You have $\displaystyle \text{}\sin^2\text{''}$ sitting by itself. You are taking the sine-squared of what?