Our instructor had given us an equation and we should get the derivative of it. But even him, he is confused what to do (funny college instructor) if we should use identities before deriving or either derive it early. Here is the equation:
$$y = \sin^2(3x)$$.
Since the answer of two ways are different. What should be the correct way?
On
Use the chain rule!
If $h(x) = f(g(x))$ then $h'(x) = f(g(x))*g'(x)$
but it appears that you will have to perform the chain rule twice!
On
Are you familiar with the chain rule?
Apply it twice and you will have that:
$$y' = 2\sin(3x)\cos(3x)3 = 6\sin(3x)cos(3x)$$
We used here that:
$f(x) = x^2 \implies f'(x) = 2x$
$g(x) = \sin(x) \implies g'(x) = cos(x)$
$h(x) = ax \implies h'(x) = a$
On
What identities are you referring to?
I will assume you are talking about the identity $\sin^2(3x) = 1 - \cos^2 (3x)$.
This says to you that $\sin^2(3x)$ is equal to $1 - \cos^2 (3x)$: so, directly deriving $\sin^2(3x)$ yelds the same result as deriving $1 - \cos^2 (3x)$.
If fact:
$$\frac{d}{dx}\sin^2(3x) = 2 \sin(3x) \cos(3x) \cdot 3 = 6 \sin(3x) \cos(3x)$$
as well as
$$\frac{d}{dx} \left( 1 - \cos^2(3x) \right) = \left[ - 2 \cos(3x) \right] \left( - \sin (3x) \right) \cdot 3 = 6 \sin(3x) \cos(3x)$$
Your instructor should always obtain this result.
As others mentioned, in both cases you must obviously apply two times the chain rule for $[f(g(x))]^2 = \sin^2(3x)$ or $[f(g(x))]^2 = \cos^2(3x)$, the first time for the 2nd power of sine (or cosine) and the second time for $3x$.
You get $y' = 6 \sin(3x) \cos(3x)$ as result using chain rule and product rule. And if you derive that equation on two ways, you will always get the same result since the function is differentiable on $\mathbb R$. If your instructor says something else he is simply wrong.