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.

First, you are taking short cuts when you are writing your equations that simply confuse things.  Also, LaTeX has \sin and \cos. Start over.
$$
\begin{aligned}
\text{Let}\quad f(x) &= \sin\frac{1}{x}.\\
\text{Let}\quad u &= \frac{1}{x}.\\
f(x) &= \sin u\\
\frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \quad\text{by the Chain rule}\\
\frac{df}{du} &= \cos u\\
\frac{du}{dx} &= -\frac{1}{x^2} \\
\frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \\
&= \cos u \cdot [-\frac{1}{x^2}] \\
&=  \cos \frac{1}{x} \cdot [-\frac{1}{x^2}]\\
&=  -\frac{1}{x^2} \cos \frac{1}{x}.
\end{aligned}
$$
By the way, you are panicking and falling into the "Universal Distributive Law of Freshmen"; you think that all operations distribute over each other, and all operations commute.  This is what has caused legions of students to replace $\sqrt{1+x^2}$ with $1 + x$.  You wrote:
$$
\cos\frac1x \cdot \frac{1}{x^2} = \cos\frac{1}{x^3}.
$$
Even though $\cos x$ looks like a product, it isn't.  Had you been using a programming language, you'd see that COS(X)*Y is not the same as COS(X*Y).  Try any of these cases with actual numbers and a pocket calculator; you'll see what I mean.
When you panic like that, always take a deep breath and start over.  And, never write a function next to its derivative and equate them.
