1. ## Trig differentation

How do you derive this equation:

y = sin ^ 3 (cosx^3)

2. This is the chain rule if the cosine is inside the sine, which what this seems to be.

3. Originally Posted by mathamatics112
How do you derive this equation:

y = sin ^ 3 (cosx^3)
You need to use the chain rule several times:

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

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

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

And so on. Can you continue?

4. 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?

5. 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?

6. i'm not seeing it

7. Originally Posted by mathamatics112
i'm not seeing it
Did you look at the part that I highlighted? You have $\text{}\sin^2\text{''}$ sitting by itself. You are taking the sine-squared of what?