Answer
y'=$\frac{-sec(\frac{1}{x})tan(\frac{1}{x})}{x^{2}}$
Work Step by Step
The chain rule states that if y=g(h(x)), then y'=g'(h(x))h'(x). Since y=sec($\frac{1}{x}$), we can set g(x)=sec(x) (outside function) and h(x)=$\frac{1}{x}$ (inside function), and use the chain rule to find y'. Therefore,
y'=g'($\frac{1}{x}$)h'(x)
We know that
$\frac{d}{dx}$[sec(x)]=sec(x)tan(x), which can be determined by using the quotient rule to compute the derivative of the function g(x)=$\frac{1}{cos(x)}$, which of course is equivalent to sec(x)
$\frac{d}{dx}$[$\frac{1}{x}$]=-1$x^{-2}$, using the power rule (rewrite $\frac{1}{x}$ as $x^{-1}$)
Now that we know that g'(x)=sec(x)tan(x) and h'(x)=-1$x^{-2}$, we can find y'.
y'=sec($\frac{1}{x}$)tan($\frac{1}{x}$)*-1$x^{-2}$
=$\frac{-sec(\frac{1}{x})tan(\frac{1}{x})}{x^{2}}$
