Determine $f'(12)$ when $f(x)=x\cos(12/x)$. Chain Rule Determine $f'(12)$ when $f(x)=x\cos(12/x)$ 
So I'm learning about chain-rule. I know that $x\cos(x)$ is the outside and $(12/x)$ is the inside.  So I did derivatives for both sides. So my function is now:
$-x\sin(12/x)\cdot\frac{d}{dx}(12/x)$
I was wondering if I did the formula above right, and do I use the quotient derivative rule for $(12/x)$? 
Please Help!!!
 A: You need both the chain rule and the product rule here!
$$\text{Given}\;\;f(x) = g(x)\cdot h(x)$$
Then $$f'(x) = g'(x)h(x) + g(x)h'(x)$$
In your case, $g(x) = x$, and $h(x) = \cos\left(\frac{12}x\right)$.
Use the chain rule to calculate $h'(x)$. While you can use the quotient rule, note that we can easily use the power rule to compute $\frac {d}{dx}\left(\frac{12}x\right) = \frac{d}{dx}\left(12x^{-1}\right)$, you can easily use the power rule. $$\begin{align}h(x) = \cos\left(\frac{12}x\right) \implies h'(x) & = -\sin \left(\frac {12}{x}\right)\frac{d}{dx}\left(\frac {12}{x}\right)\\ \\ & = -\sin \left(\frac {12}{x}\right)\frac d{dx}(12x^{-1}) \\ \\ & = -\sin \left(\frac {12}{x}\right)\left(\frac {-12}{x^2}\right) \\ \\ & = \frac{12\sin\left(\frac{12}{x}\right)}{x^2}\end{align}$$
A: You missed the initial product rule part:
$$f'(x) = x\frac{d}{dx}\sin\left(\frac{12}{x}\right) + \sin\left(\frac{12}{x}\right)\frac{d}{dx}x$$
And you may use quotient rule for $\frac{d}{dx}\frac{12}{x}$, but you may also use power rule after removing the constant $12$. Try both, and they should give the same answer.
A: basically 
f'(x) = x[ $ - sin(12/x)$ * $(-1/x^2) $ + $ cos(12/x)(1)$.
now put x = 12  and your job is done
