Derivative of Trig Functions (Intuition Help?) Looking for some intuition help here.
I have the following exercise and these are the steps I take:
$$
y = \sin\left(\frac{1}{x}\right)
$$
$$
u=\frac{1}{x}
$$
$$
y = \sin u,\;\;\frac{dy}{du} = \cos u= \cos\left(\frac{1}{x}\right)
$$
$$
u=x^{-1};\;\frac{du}{dx} =-x^{-2}=-\frac{1}{x^2}
$$
$$
\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=cos\frac{1}{x}\times-\frac{1}{x^2}
$$
This is incorrect but intuitively I want to multiply it this way.
$$
cos\frac{1}{x}\times-\frac{1}{x^2}=cos-\frac{1}{x^3}
$$
But the correct answer is:
$$
-\frac{cos-\frac{1}{x}}{x^2}
$$
Help me absorb the why so I can intuitively solve problems like these.

Here is where parentheses come in handy:
You found, correctly, $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$.
But the scope of $\cos$ is restricted to its argument: $\left(\dfrac 1x\right)$ ONLY: 
The FUNCTION $\dfrac{dy}{du} = \cos\left(\dfrac 1x\right)$ is multiplied by the function $\dfrac{du}{dx} = -\dfrac 1{x^2}$. That is not what you did. You multiplied argument of the $\cos$ function by the function $\dfrac{du}{dx} = -\dfrac{1}{x^2}$.
$$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\left[cos\left(\frac{1}{x}\right)\right]\times\left(-\frac{1}{x^2}\right) $$ $$= -\frac 1{x^2} \cos\left(\frac 1x\right)
= -\dfrac{\cos\left(\frac 1x\right)}{x^2}$$
