Trig Differentaition 
Differentiate $y=27 \sec^3(x)$ with respect to $x$.

I tried splitting the $\sec^3(x)$ into $\sec^2(x)\cdot \sec(x)$ and using the product rule but that didn't work.
 A: $$y=27 \sec^3(x)$$
$$y=27 \times(\sec (x))^3$$
Now apply the chain rule...
The "outside" function is the function 'to cube'
The "inside" function is the $\sec(x)$
So focusing on the cube on the "outside"...
$$y=27 \times (inside)^3$$
$$\frac{dy}{dx}=27\times 3 \times (inside)^2 \times (derivative-of- inside)$$
then focussing on the "inside"...
$$\frac{dy}{dx}=27\times 3 \times (\sec(x))^2 \times (derivative-of- \sec(x))$$
$$\frac{dy}{dx}=27\times 3 \times (\sec(x))^2 \times (\sec(x)\tan(x))$$
$$\frac{dy}{dx}=81\times (\sec(x))^3 \times \tan(x)$$
$$\frac{dy}{dx}=81 \sec^3(x) \tan(x)$$
A: If $y = 27\sec^{3}(x)$, then you can find $y'$ in a couple of ways.
Using the method you suggested:
$$y = 27\sec^{3}(x) = 27\sec^{2}(x)\cdot\sec(x)$$
Applying the product rule, we have:
$$y' = 27\dfrac{d}{dx}\big[\sec^{2}(x)\cdot\sec(x) \big] = 27\bigg(\dfrac{d}{dx}\big[\sec^{2}(x)\big]\cdot \sec(x) + \sec^{2}(x)\dfrac{d}{dx}\big[\sec(x)\big]\bigg)$$
One would need to find $\dfrac{d}{dx}\sec^{2}(x)$ by the chain rule (or by applying the product rule again).
By the chain rule:
$$y = (f\circ g)(x) = f(g(x)),\quad \text{ where } g(x) = \sec(x),\quad f(x) = 27x^3$$
Then $$y' = f'(g(x))\cdot g'(x) = 3\cdot (\sec(x))^2 \cdot g'(x)$$
Does this make sense?
A: Rewrite $y=z^3$ where $z=3\sec x$. As $z$ is a function of $x$ $$\dfrac{dy}{dx}=3z^2\cdot \dfrac{dz}{dx}\;\text{with}\;\dfrac{dz}{dx}=3\sec x\tan x$$ So $$\dfrac{dy}{dx}=27\sec^2 x\cdot(3 \sec x\tan x)$$
